Electricity and Magnetism Experiments

Last week, my middle schoolers did a set of experiments on electricity and magnetism. They answered the questions:

How does the voltage across each light bulb change as you add more and more bulbs to a parallel circuit?
How does the voltage across each light bulb change as you add more and more bulbs to a series circuit?
How does the number of coils of wire wrapped around a nail affect it’s magnetism (as measured by the number of paperclips it can pick up)?
How does the amount of salt mixed into water affect its conductivity?

An electromagnetic nail lifts two paperclips.

Students measure the conductivity of a salt water solution.

Each question is designed so that students have something to measure and will be able to use those measurements to make predictions. For example, once they’ve measured the voltage across four bulbs in series, they should be able to predict the voltage across the bulbs in a series of ten.

Some of the experiments, like the nail electromagnet, should have simple linear trends, with students choosing the advanced option having to find an equation to fit their data for the predictions. And I’ll challenge the students in Algebra II to find the equations for the inverse relationships–I’ve already asked their math teacher (Mr. Schmidt) to help them out if they need it.

This has also provided the opportunity for them to apply what they’ve just learned about drawing circuit diagrams (we use this set of symbols).

Circuit diagrams of bulbs in parallel. The voltage difference across each bulb is also noted.

Experiments for Demonstrating Different Types of Mathematical Functions

This is my quick, and expanding, reference for easy-to-do experiments for students studying different types of functions.

Linear equations: y = mx + b

Bringing water to a boil (e.g. Melting snow)
Straight line, motorized, motion. (e.g. Movement of a robot/Predicting where robots will meet in the middle)
Current versus Voltage across a resistor as resistance changes.

Quadratic equations: y = ax² + bx + c

Ballistics: throwing a ball (e.g. soccer ball ballistics)

Exponential functions: y = ae^kx

Cooling water (ref.)

Square Root Functions: y = ax^1/2

Draining water from a container: (e.g. Draining water)

Trigonometric Functions: y = asin(bx)+c

Analyzing the Motion of Soccer Ball using a Camera and Calculus

projectile-3832c2-800 — Animation showing the motion of the ballistic motion of a soccer ball.

If you throw a soccer ball up into the air and take a quick series of photographs you can capture the motion of the ball over time. The height of the ball can be measured off the photographs, which can then be used for some interesting physics and mathematics analysis. This assignment focuses on the analysis. It starts with the height of the ball and the time between each photograph already measured (Figure 1 and Table 1).

projectile-diagram-plain — Figure 1. Height of a thrown ball, measured off a series of photographs. The photographs have been overlaid to create this image of multiple balls.

Table 1: Height of a thrown soccer ball over a period of approximately 2.5 seconds. This data were taken from a previous experiment on projectile motion.

Photo	Time (s)	Measured Height (m)
P₀	0	1.25
P₁	0.436396062	6.526305882
P₂	0.849230104	9.825317647
P₃	1.262064145	11.40310588
P₄	1.674898187	11.30748235
P₅	2.087732229	9.657976471
P₆	2.50056627	6.191623529

Assignment

Pre-Algebra: Draw a graph showing the height of the ball (y-axis) versus time (x-axis).
Algebra/Pre-calculus: Determine the equation that describes the height of the ball over time: h(t). Plot it on a graph.
Calculus: Determine the equation that shows how the velocity of the ball changes over time: v(t).
Calculus: Determine the equation that shows how the acceleration of the ball changes with time: a(t)
Physics: What does this all mean?

Endothermic Reactions: Vinegar and Baking Soda

A quick and simple experiment that demonstrates endothermic reaction and can include a discussion of ionic and covalent bonds. Mixing baking soda and vinegar together drops the temperature of the liquid by about 4 °C in one minute. (Note that while the temperature drops and the reaction looks endothermic, it’s actually not — other things cause the cooling. However, since it looks like an endothermic reaction I use it as a first approximation of one.)

Ingredients

3 g baking soda – (sodium bicarbonate – NaHCO₃)
60 ml vinegar – (acetic acid – CH₃COOH)
200 ml styrofoam cup (needs to be big enough to contain the bubbles).
thermometer

Procedure

Add the baking soda to the vinegar in the styrofoam cup. Measure the temperature while stirring for about a minute.

Results

Time (t)	Temperature (°C)
0	25
15	24
30	21
60	21

Discussion

The chemical reaction between baking soda (sodium bicarbonate) and vinegar (acetic acid) can be written:

NaHCO₃ + CH₃COOH —-> CO₂ + H₂O + CH₃OONa

The products of the reaction are carbon dioxide gas (which gives the bubbles), water, and sodium acetate.

However, a more detailed look shows that for the reaction to work the two chemicals need to be dissolved in water. Dissolving these ionic compounds causes the two ions to separate. Dissolved baking soda dissociates into a sodium and a bicarbonate ion:

sodium bicarbonate —-> sodium ion + bicarbonate ion

NaHCO₃ —-> Na⁺ + HCO₃^–

Why doesn’t the bicarbonate break into smaller pieces? Because it’s atoms are bonded together more tightly by covalent bonds.

Similarly, the acetic acid in vinegar dissociates into:

acetic acid —-> hydrogen ion + acetate

CH₃COOH —-> H⁺ + CH₃COO^–

This video has a nice overview of ionic versus covalent bonding.

References

More detail about the reaction can be found at:

Wayne (accessed 2012): Baking Soda and Vinegar Reaction and Demonstrations, on http://apple-cider-vinegar-benefits.com

Draining a Bottle Part 2: Linearizing Equations when you have to

Yesterday we used calculus to find the equation for the height of water in a large plastic water bottle as the water drained out of a small hole in the bottom.

Perhaps the most crucial point in the procedure was fitting a curve to the measured reduction of the water’s outflow rate over time. Yesterday, in our initial attempt, we used a straight line for the curve, which produced a very good fit.

q-v-t — Figure 1. The change in the outflow rate over time can be well approximated by a straight line.

The R² value is a measure of how good a fit the data is to the trendline. The straight line gives an R² value of 0.9854, which is very close to a perfect fit of 1.0 (the lowest R² can go is 0.0).

The resulting equation, written in terms of the outflow rate (dV/dt) and time (t), was:

$\frac{dV}{dt} = -0.0035 t + 3.9113$

However, if you look carefully at the graph in Figure 1, the last few data points suggest that the outflow does not just linearly decrease to zero, but approaches zero asymptotically. As a result, a different type of curve might be a better trendline.

Types of Equations

So my calculus students and I, with a little help from the pre-Calculus class, tried to figure out what types of curves might work. There are quite a few, but we settled for looking at three: a logarithmic function, a reciprocal function, and a square root function. These are shown in Figure 2.

type-curves-300 — Figure 2. Example curves that might better describe the relationship between outflow and time.

I steered them toward the square root function because then we’d end up with something akin to Torricelli’s Law (which can be derived from the physics). A basic square root function for outflow would look something like this:

$\frac{dV}{dt} = a \sqrt{t} + b$

the a coefficient stretches the equation out, while the b coefficient moves the curve up and down.

Fitting the Curve

Having decided on a square-root type function, the next problem was trying to find the actual equation. Previously, we used Excel to find the best fit straight line. However, while Excel can fit log, exponential and power curves, there’s no option for fitting a square-root function to a graph.

To get around this we linearized the square-root function. The equation, after all, looks a lot like the equation of a straight line already, the only difference is the square root of t, so let’s substitute in:

$x = \sqrt{t}$

to get:

$\frac{dV}{dt} = a x + b$

Now we can get Excel to fit a straight line to our data, but we have to plot the square-root of time versus temperature instead of the just time versus temperature. So we take the square root of all of our time measurements:

Time	Square root of time	Outflow rate
t (s)	t^1/2 = x (s^1/2)	dV/dt (ml/s)
0.0	0	3.91
45.5	6.75	3.52
97.8	9.89	2.94
140.9	11.87	3.52
197	14.04	3.21
257	16.05	3.01
315.1	17.75	2.81
380.1	19.50	2.53
452.9	21.28	2.23
529.6	23.01	1.92
620.7	24.91	1.69
742.7	27.25	1.45

We can now plot the outflow rate versus the square root of time (Figure 3).

q-v-sqrt(t) — Figure 3. Linear trend relating the outflow rate to the square root of time. The regression coefficient (R²) of 0.9948 is better than the simply linear trend of outlfow rate versus time (which was 0.9854).

The equation Excel gives (Figure 3), is:

$\frac{dV}{dt} = -0.1395 x + 5.21$

and we can substitute back in for x=t^1/2 to get:

$\frac{dV}{dt} = -0.1395 \sqrt{t} + 5.21$

Getting back to the Equation for Height

Now we can do the same procedure we did before to find the equation for height.

First we substitute in V=πr²h:

$\frac{d(\pi r^2 h}{dt} = -0.1395 \sqrt{t} + 5.21$

Factor out the πr² and move it to the other side of the equation to solve for the rate of change of height:

$\frac{dh}{dt} = \frac{-0.1395}{\pi r^2} \sqrt{t} + \frac{5.21}{\pi r^2}$

Then integrate to find h(t) (remember $\sqrt{t} = t^{1/2}$ ) :

$\int \frac{dh}{dt} dt = \int \left( \frac{-0.1395}{\pi r^2} t^{1/2} + \frac{5.21}{\pi r^2} \right) dt$

gives:

$h = \frac{-0.1395}{(3/2) \pi r^2} t^{3/2} + \frac{5.21}{\pi r^2} t + c$

which might look a bit ugly, but that’s only because I haven’t simplified the fractions. Since the radius (r) is 7.5 cm:

$h = -0.000526 t^{3/2} + 0.029 t + c$

Finally we substitute in the initial value (t=0, h=11) to solve for the coefficient:

$c = 11$

giving the equation:

$h = -0.000526 t^{3/2} + 0.029 t + 11$

Plotting the equations shows that it matches the measured data fairly well, although not quite as well as when we used the previous linear function for outflow.

h-v-t_sqrt — Figure 4. Integrating a square root function for the outflow rate gives a modeled function for the changing height over time that slightly undermatches the measured heights.

Discussion

I’m not sure why the square root function for outflow does not give as good a match of the measurements of height as does the linear function, especially since the former better matches the data (it has a better R² value).

It could be because of the error in the measurements; the gradations on the water bottle were drawn by hand with a sharpie so the error in the height measurements there alone was probably on the order of 2-3 mm. The measurement of the outflow volume in the beaker was also probably off by about 5%.

I suspect, however, that the relatively short time for the experiment (about 15 minutes) may have a large role in determining which model fit better. If we’d run the experiment for longer, so students could measure the long tail as the water height in the bottle got close to the outlet level and the outflow rate really slowed down, then we’d have found a much better match using the square-root function. The linear match of the outflow data produces a quadratic equation when you integrate it. Quadratic equations will drop to a minimum and then rise again, unlike the square-root function which will just continue to sink.

Conclusions

The linearization of the square-root function worked very nicely. It was a great mathematical example even if it did not produce the better result, it was still close enough to be worth it.

Watching Snow Melt: Observing Phase Changes and Latent Heat

snow-0638 — Waiting, observing, and recording as the snow melts on the hot plate.

Though it might not sound much more interesting than watching paint dry, the relationships between phase changes, heat, and temperature are nicely illustrated by melting a beaker of snow on a hot plate.

snow-river-4205-s — A light, overnight snowfall, lingers on the branches that cross the creek.

This week’s snowfall created an opportunity I was eager to take. We have access to an ice machine, but closely packed snow works much better for this experiment, I think; the small snowflakes have larger surface-area to volume ratio, so they melt much more evenly, and demonstrate the latent heat of melting much more effectively.

Instructions

My instructions to the students are simple: collect some snow, and observe how it melts on the hot plate.

I also ask them to determine the mass and density of the snow before (and after) the melting, so I could show that throughout the phase changes and transformations the mass does not change (at least not a lot) and so they can practice calculating density^1,2.

Procedure

I broke up my middle school students into groups of 2 or 3 and had them come up with a procedure and list of materials before they started. As usual I had to restrain a few of the over-eager ones who wanted to just rush out and collect the snow.

snow-0637 — A 600 ml beaker filled with (cold) snow. A thermometer is embedded in the ice.

I guided their decision-making a little, so they would use glass beakers for the collection and melting. Because I wasn’t sure what the density of the packed snow would be, I suggested the larger, 600 ml beakers, which turned out to work very well. They ended up with somewhere between 350 and 400 grams of snow, giving densities around 0.65 g/ml.

When they put the beakers on the hot-plate, I specifically asked the students to observe and record, every minute or so, the changes in:

temperature,
volume
appearance

I had them continue to record until the water was boiling. This produced the question, “How do we know when it’s boiling?” My answer was that they’d know when they saw the temperature stop changing.

They also needed to stir the water well, especially when the ice was melting, so they could get a “good”, uniform temperature reading.

Results

We ended up with some very beautiful graphs.

Temperature Change

snow-T_data-0643 — Changing temperature with time as the beaker of snow melted into water and then came to a boil. Graph by E.F.

The temperature graph clearly shows three distinct segments:

In the first few minutes (about 8 min), the temperature remains relatively constant, near the freezing/melting point of water: 0 ºC.
Then the temperature starts to rise, at an constant rate, for about 20 minutes.
Finally, when the water reaches close to 100 ºC, its boiling point, the temperature stops changing.

Volume Change

The graph of volume versus time is a little rougher because the gradations on the 600 ml beaker were about 25 ml apart. However, it shows quite clearly that the volume of the container decreases for the first 10 minutes or so as the ice melts, then remains constant for the rest of the time.

snow-V_data-0642 — The change in volume with time of the melting ice. Graph by E.F.

Analysis

To highlight the significant changes I made copies of the temperature and volume graphs on transparencies so they could be overlain, and shown on the overhead projector.

Melting Ice: Latent Heat of Melting/Fusion

snow-melt — Comparison of temperature and volume change data shows that the temperature starts to rise when the volume stops changing.

The fact that the temperature only starts to rise when the volume stops changing is no coincidence. The density of the snow is only about 65% of the density of water (0.65 g/ml versus 1 g/ml), so as the snow melts into water (a phase change) the volume in the beaker reduces.

When the snow is melted the volume stops changing and then the temperature starts to rise.

The temperature does not rise until the snow has melted because during the melting the heat from the hot plate is being used to melt the snow. The transformation from solid ice to liquid water is called a phase change, and this particular phase change requires heat. The heat required to melt one gram of ice is called the latent heat of melting, which is about 80 calories (334 J/g) for water.

Conversely, the heat that needs to be taken away to freeze one gram of water into ice (called the latent heat of fusion) is also 80 calories.

So if we had 400 grams of snow then, to melt all the ice, it would take:

400 g × 80 cal/g = 32,000 calories

Since the graph shows that it takes approximately 10 minutes (600 seconds) to melt all the snow the we can calculate that the rate at which heat was added to the beaker is:

32,000 cal ÷ 10 min = 3,200 cal/min

Constantly Rising Temperature

The second section of the temperature graph, when the temperature rises at an almost constant rate, occurs after all the now has melted and the beaker is now full of water. I asked my students to use their observations from the experiment to annotate the graphs. I also asked a few of my students who have worked on the equation of a line in algebra to draw their best-fit straight lines and then determine the equation.

snow-graph-anno-0644 — The rising temperatures in the middle of the graph can be modeled with a straight line. Graph by A.F.

All the equations were different because each small group started with different masses of snow, we used two different hot plates, and even students who used the same data would, naturally, draw slightly different best-fit lines. However, for an example, the equation determined from the data shown in the figure above is:

y = 4.375 x – 35

Since our graph is of Temperature (T) versus time (t) we should really write the equation as:

T = 4.375 t – 35

It is important to realize that the slope of the line (4.375) is the change in temperature with time, so it has units of ºC/min:

slope = 4.375 ºC/min

which means that the temperature of the water rises by 4.375 ºC every minute.

NOTE: It would be very nice to be able to have all the students compare all their data. Because of the different initial masses of water we’d only be able to compare the slopes of the lines (4.375 ºC/min in this case, but another student in the same group came up with 5 ºC/min).

Furthermore, we would also have to normalize with respect to the mass of the ice by dividing the slope by mass, which, for the case where the slope was 4.375 ºC/min and the mass was 400 g, would give:

4.375 ºC/min ÷ 400 g = 0.011 ºC/min/g

Specific Heat Capacity of Water

A better alternative for comparison would be to figure out how much heat it takes to raise the temperature of one gram of water by one degree Celsius. This can be done because we earlier calculated how much heat is being added to the beaker when we were looking at the melting of the ice.

In this case, using the heating rate of 3,200 cal/min, a mass of 400 g, and a rising temperature rate (slope from the curve) of 4.375 ºC/min we can:

3,200 cal/min ÷ 4.375 ºC/min ÷ 400 g = 1.8 cal/ºC/g

The amount of heat it takes to raise the temperature of one gram of a substance by one degree Celsius is called its specific heat capacity. We calculated a specific heat capacity of water here of 1.8 cal/ºC/g. The actual specific heat capacity of water is 1 cal/ºC/g, so our measurements are a wee bit off, but at least in the same ballpark (order of magnitude). Using the students actual mass measurements (instead of using the approximate 400g) might help.

Evaporating Water

Finally, in the last segment of the graph, the temperature levels off again at about 100 ºC when the water starts to boil. Just like the first part where the ice was melting into water, here the water is boiling off to create water vapor, which is also a phase change and also requires energy.

The energy required to boil one gram of water is 540 calories, which is called the latent heat of vaporization. The water will probably remain at 100 ºC until all the water boils off and then it will begin to rise again.

Conclusion

This project worked out very well, and there was so much to tie into it, including: physics, algebra, and graphing.

Notes

¹ Liz LaRosa (2008) has a very nice density demonstration comparing a can of coke to one of diet coke.

² You can find the density of most of the elements on the periodic table at periodictable.com.

Projectile Motion

Abstract

A series of still photographs of a projectile (soccer ball) in motion were used to determine the equation for the height of the ball (h(t) = 4.9 t² + 14.2 t + 1.25), the initial velocity of the ball (14.2 m), the maximum height of the ball (11.6 m), and the time between each photograph (0.41 s). The problem was solved numerically using MS Excel’s Solver function. There are much easier ways of doing this, which we did not do.

Introduction

projectile-diagram — Figure 1. Calculated elevation of the soccer ball after launch.

One of physics lab assignments I gave my students was to see if students could use a camera to capture a sequence of images of a projectile, plot the elevation of the projectile from the photographs, determine the constants in the parabolic equation for the height of the projectile, and, in so doing, determine the velocity at which the projectile was launched.

I offered my old, digital Pentax SLR that can take up to seven pictures in quick sequence and be set to fully manual. A digital video camera with a detailed timestamp would have been ideal, but we did not have one available at the time.

Now the easy way of getting the velocity data would be to estimate the heights (h) of the ball from the image using some sort of known reference (in this case the whiteboard), and determine the time between each photograph (Δt) by photographing a stopwatch using the same shutterspeed settings. After all, the average velocity of the ball between two images would be:

$! \bar{v} = \frac{\Delta h}{\Delta t}$

The reference whiteboard is four feet tall (1.22 m) in real life, but 51 pixels tall in the image. Using this ratio (i.e. 1.22 m = 51 px) we can convert the heights of the ball from pixels to meters:

elevations — Table 1. Table showing the conversion of the height of the ball in pixels to elevation (in meters).

Unfortunately, I think my students forgot to do the pictures of the stopwatch to get Δt, the time between each photograph. Since the lab reports are due on Monday, and it’s the weekend now I’m curious to see what they come up with.

However, I was wondering if they could use just the elevation data to back out the Δt. So I gave it a try myself. Even the easiest way of solving this problem is not trivial, in fact, I ended up resorting to Excel’s iterative solver to find the answers. While this procedure probably goes a little beyond what I expect from the typical high school physics student, more advanced students who are taking calculus might benefit.

Procedure

We took the reference whiteboard (1.21 m tall), a soccer ball, and the camera outside. The whiteboard was leant vertically against the post of the soccer goal. The ball was thrown vertically by a student standing next to the whiteboard (see Figure 1) while pictures were taken. The camera’s shutterspeed was 1/250th of a second. The distance from the camera to the person throwing the ball (and to the whiteboard) were not measured.

The procedure was repeated several times, but only one trail was used in this analysis.

The images were loaded onto a computer, and the program GIMP was used to determine the distance, in pixels, from the ground to the projectile. The size of the reference whiteboard, in pixels, was used to calculate the height of the soccer ball in meters.

The elevations measured off the photographs were then used to calculate the release velocity, time between snapshots, and maximum height of the ball.

The Equation for Elevation

I started with the fact that once the ball is released, the only force acting on it is the force of gravity. Since the mass of the ball does not change we only have to consider the acceleration due to gravity (-9.8 m/s²). I also neglect air resistance to make things easier.

Finding the Velocity Equation

Start with the fact that, acceleration is the rate of change of velocity with time. You can write it in the differential form:

$! a = \frac{dv}{dt} = -9.8$

so we integrate with respect to time to get the equation for velocity as a function of time:

$! v(t) = \int -9.8 dt$

$! v(t) = -9.8 t + c$

where c is an unknown constant. What we do know though, is that at the beginning, when the ball is just launched, time is zero (t = 0) so c_v becomes the initial velocity (v₀) at which the ball is thrown:

at t = 0, v(0) = v₀:
$! v_0 = -9.8 (0) + c$

$! v_0 = c$

So our velocity equation becomes:

$! v(t) = -9.8 t + v_0$

Finding the height equation

Now since we know that velocity is the rate of change of distance (in this case height) with time:

$! v(t) = \frac{dh}{dt} = -9.8 t + v_0$

so we integrate again to find the height equation:

$! \frac{dh}{dt} = -9.8 t + v_0$
$! h(t) = \int (-9.8 t + v_0) dt$
$! h(t) = \frac{-9.8 t^2}{2} + v_0 t + c$

Similar to what we did with the velocity equation, to find the new constant c we consider what happens at the start time, when the ball is launched, and t = 0 and h(0) = h₀;

$! h_0 = \frac{-9.8 (0)^2}{2} + v_0 (0) + c$

so:
$! h_0 = c$

The constant is equal to the initial height of the ball — the height of the ball when it’s thrown. So we end up with the final equation:

$! h(t) = \frac{-9.8 t^2}{2} + v_0 t + h_0$

Results

Solving all the unknowns

At this point, although we have an equation for the height of the ball, we don’t know the initial velocity (v₀), nor do we know the initial height of the ball when it’s released (h₀). And we still don’t know the time when the ball is at each position.

With that many unknowns we’d need the same number of independent equations to be able to solve for them all. It may be possible, but instead of analytically solving the equations I opted to take a numerical approach, and use Excel’s Solver function.

I started by setting up the equations to calculate the height of the ball at six different times to correspond with our six height measurements. It was necessary therefore to create a set of variables:

Time when we started taking pictures (t₁): Since we don’t know how long after we threw the ball we started taking pictures, I made this a variable called t₁.
The time between each picture (dt): I made the assumption that the time between each picture would be constant. The shutter speed was constant (1/250th of a second) so there is no obvious reason why the time should be different.
Initial velocity (v₀): The initial upward speed at which the ball was thrown. Obviously, the faster the initial speed the higher the ball goes, so this is a fairly important parameter.
Initial height (h₀): We also don’t precisely know how high the ball was when it was released, so this also needs to be a variable.

By defining the time between each picture as dt, we can write the time that each picture was taken in terms of the time of the initial picture (t₁) and dt. After all the second picture would have been taken dt seconds after the first for a total time of:

$! t(P2) = t_0 + dt$

similarly for all the pictures:

photo-times — Table 2. Table of expressions giving the time when each of the six photos were taken.

Now I set up an Excel spreadsheet and gave all the unknown variable values and initial value of 1:

excel-solver — Table 3. Table in Excel for determining the height of a projectile. All of the unknown variables' values are highlighted in green and have been given an initial value of 1.

Now I just had to run Solver and tell it that I wanted the Total Residual, which gives the difference between the h(1) equation’s values for height and the actual, measured values, to be as close to zero as possible. A perfect fit of the equation to the data would have a total residual of one, but that’s not possible when you’re dealing with real data.

solving — Table 4. Parameters set in Solver to determine the values of the unknown constants.

Even so, I had to goose Solver a bit for it to produce reasonable numbers. I put in a few constraints:

dt >= 0: We could not have a negative time between pictures.
h₀ <= 1.25: 1.25 meters seemed reasonable for the height at which the ball was released.
t₁ <= 1: It also seemed reasonable that the time when the first picture was taken was less than one second after the ball was thrown.

I ran the Solver a few times, and had to reset dt to 0.5 at one point when it had become zero, but the final result looked remarkably good: the total difference between the modeled line and the actual data was only 0.113 meters.

solver-after — Table 5. Solver's solution for the unknown constants in the height equation.

So we found that:

Initial velocity: v₀ = 14.2 m/s
Height at release: h₀ = 1.25 m
Time between pictures: dt = 0.41 s
Time when the first picture was taken: t₁ = 0.44 s

Which makes the height equation:

$! h(t) = \frac{-9.8 t^2}{2} + 14.2 t + 1.25$

Using these constants in the height equation, we could see how good fit the height equation was to the data:

projectile-graph — Figure 3. Graph comparing the modeled heights (from the h(t) equation) to the actual data.

Maximum Height of the Ball

Finally, the maximum height of the ball can be read off the graph, but it can also be determined using the equation for the height of the ball:

$! h(t) = \frac{-9.8 t^2}{2} + v_0 t + h_0$

We know that the maximum height is reached when the ball stops moving upward and starts to descend. At that point, the vertical velocity of the ball is zero. Since the velocity of the ball is the rate of change of height ( $v = \frac{dh}{dt}$ ) we can differentiate the height equation to get an equation for velocity.

$! h(t) = \frac{-9.8 t^2}{2} + v_0 t + h_0$

$! v = \frac{dh}{dt} = -9.8 t + v_0$

since we’ve determined that the initial velocity of the ball is 14.2 m/s we get:

$! v = -9.8 t + 14.2$

when the velocity is zero (v = 0):

$! 0 = -9.8 t + 14.2$

which can be solved for t to find that the time the ball reaches it’s maximum height is:

$! t = 1.45 seconds$

Putting this into the height equation:

$! h(1.45) = \frac{-9.8 (1.45)^2}{2} + 14.2 (1.45) + 1.25$

gives:

$! h_{max} = 11.58 meters$

Discussion

I’m quite happy with the way this project turned out. The fit between the modeled heights (h(t)) and the actual heights was very good.

My primary concern going into the project was that the distortion from the camera lens would make this technique impossible, but that appears not to be a significant problem.

Most of this calculation, including the somewhat tricky numerical solution using Solver could have been avoided if I’d calibrated the camera, simply by pointing it at a stopwatch (using the same shutterspeed as in the experiment) and measuring the time between snapshots. It will therefore be interesting to see if the actual time between shots (dt) is close to the dt of 0.41 seconds calculated by the model.

Finally, as noted above, a video camera with a timestamp would possibly be a more useful technology for this experiment.

Conclusion

It is possible to analyze the projectile path of an object using a series of snapshots, to determine the initial velocity of the projectile, its release height, and the time between snapshots, if you can assume that the time between snapshots is identical. There are, however, much easier methods of solving this problem.

References

None, but this is where they’d be if I had any.

Appendix

The Excel spreadsheet where all the calculations were done is here.

Figuring Out Experimental Error

Using stopwatches, we measured the time it took for the tennis ball to fall 5.3 meters. Some of the individual measurements were off by over 30%, but the average time measured was only off by 7%.

I did a little exercise at the start of my high-school physics class today that introduced different types of experimental error. We’re starting the second quarter now and it’s time for their lab reports to including more discussion about potential sources of error, how they might fix some of them, and what they might mean.

One of the stairwells just outside the physics classroom wraps around nicely, so students could stand on the steps and, using stopwatches, time it as I dropped a tennis ball 5.3 meters, from the top banister to the floor below.

measured-falling-time — Students' measured falling times (in seconds).

Random and Reading Errors

They had a variety of stopwatches, including a number of phones, at least one wristwatch, and a few of the classroom stopwatches that I had on hand. Some devices could do readings to one hundredth of a second, while others could only do tenths of a second. So you can see that there is some error just due to how detailed the measuring device can be read. We’ll call this the reading error. If the best value your stopwatch gives you is to the tenth of a second, then you have a reading error of plus or minus 0.1 seconds (±0.1 s). And you can’t do much about this other than get a better measuring device.

Another source of error is just due to random differences that will happen with every experimental trial. Maybe you were just a fraction of a second slower stopping your watch this time compared to the last. Maybe a slight gust of air slowed the balls fall when it dropped this time. This type of error is usually just called random error, and can only be reduced by taking more and more measurements.

Our combination of reading and random errors, meant that we had quite a wide range of results – ranging from a minimum time of 0.7 seconds, to a maximum of 1.2 seconds.

So what was the right answer?

Well, you can calculate the falling time if you know the distance (d) the ball fell (d = 5.3 m), and its acceleration due to gravity (g = 9.8 m/s²) using the equation:

$! t = \sqrt{\frac{2d}{g}}$

which gives:

$! t = 1.043 s$

So while some individual measurements were off by over 30%, the average value was off by only 8%, which is a nice illustration of the phenomenon that the more measurements you take, the better your result. In fact, you can plot the improvement in the data by drawing a graph of how the average of the measurements improves with the number of measurements (n) you take.

falling-averages — The first measurement (1.2 s) is much higher than the calculated value, but when you incorporate the next four values in the average it undershoots the actual (calculated) value. However, as you add more and more data points into the average the measured value gets slowly closer to the calculated value.

More measurements reduce the random error, but you tend to get to a point of diminishing returns when you average just does not improve enough to make it worth the effort of taking more measurements. The graph shows the average slowly ramping up after you use five measurements. While there are statistical techniques that can help you determine how many samples are enough, you ultimately have to base you decision on how accurate you want to be and how much time and energy you want to spend on the project. Given the large range of values we have in this example, I would not want to use less than six measurements.

Systematic Error

But, as you can see from the graph, even with over a dozen measurements, the average measured value remains persistently lower than the calculated value. Why?

This is quite likely due to some systematic error in our experiment – an error you make every time you do the experiment. Systematic errors are the most interesting type of errors because they tell you that something in the way you’ve designed your experiment is faulty.

The most exciting type of systematic error would, in my opinion, be one caused by a fundamental error in your assumptions, because they challenge you to fundamentally reevaluate what you’re doing. The scientists who recently reported seeing particles moving faster than light made their discovery because there was a systematic error in their measurements – an error that may result in the rewriting of the laws of physics.

In our experiment, I calculated the time the tennis ball took to fall using the gravitational acceleration at the surface of the Earth (9.8 m/s²). One important force that I did not consider in the calculation was air resistance. Air resistance would slow down the ball every single time it was dropped. It would be a systematic error. In fact, we could use the error that shows up to actually calculate the force of the air resistance.

However, since air resistance would slow the ball down, it would take longer to hit the floor. Unfortunately, our measurements were shorter than the calculated falling time so air resistance is unlikely to explain our error. So we’re left with some error in how the experiment was done. And quite frankly, I’m not really sure what it is. I suspect it has to do with student’s reaction times – it probably took them longer to start their stopwatches when I dropped the ball than it did to stop them when the ball hit the floor – but I’m not sure. We’ll need further experiments to figure this one out.

In Conclusion

On reflection, I think I probably would have done better using a less dense ball, perhaps a styrofoam ball, that would be more affected by air resistance, so I can show how systematic errors can be useful.

Fortunately (sort of) in my demonstration I made an error in calculating the falling rate – I forgot to include the 2 under the square root sign – so I ended up with a much lower predicted falling time for the ball – which allowed me to go through a whole exercise showing the class how to use Excel’s Goal Seek function to figure out the deceleration due to air resistance.

My Excel Spreadsheet with all the data and calculations is included here.

There are quite a number of other things that I did not get into since I was trying to keep this exercise short (less than half an hour), but one key one would be using significant figures.

There are a number of good, but technical websites dealing with error analysis including this, this and this.