9.9 - Acceleration of a Falling Object
The experiment
Aim
To find the acceleration of a falling tennis ball, and to compare it with the accepted value of the gravitational field strength.
Background
- Anything dropped near the Earth's surface speeds up as it falls, pulled down by its weight. With nothing else acting, everything would gain speed at the same rate, , about 10 m/s every second, whatever its mass.
- Air resistance grows as an object speeds up, which reduces the resultant force and the acceleration. A tennis ball dropped from a metre stays slow enough to be close to free fall.
- For an object that starts from rest, the speed v after falling a distance s is given by , where is the acceleration.
- You can't measure the acceleration directly, but a light gate can measure the ball's speed just before it lands.
- So in this practical you measure the ball's speed after falling from different heights, and use a graph of speed squared against height to find the acceleration.
Variables
- Independent variable - The height the ball is dropped from.
- Dependent variable - The speed of the ball just before it lands.
- Control variables - The same tennis ball, the same light gate in the same position, and the ball released rather than thrown every time.
Method
- Set up the scale - Clamp two metre rules vertically, one above the other with their ends touching.
- Set up the light gate - Clamp it just above the floor, and connect it to the datalogger.
- Enter the diameter - Measure the diameter of the tennis ball and enter it into the datalogger.
- Drop the ball - Line the ball up over the gate with its lowest point level with the height mark, and release it without throwing.
- Record the speed - Read it from the datalogger, and repeat any drop that gives no reading or an obviously wrong one.
- Repeat at other heights - Drop the ball three times from each of five or so heights.
Safety
- Clamp the stands securely - So they can't topple, and keep feet clear of the landing area.
- Watch the bounce - A tennis ball bounces high.
Worked example - Acceleration of a falling ball
A student dropped a tennis ball through a light gate three times from each of five heights, and recorded the speed just before it landed each time.
| Height (m) | Trial 1 (m/s) | Trial 2 (m/s) | Trial 3 (m/s) |
|---|---|---|---|
| 0.20 | 1.96 | 2.00 | 1.98 |
| 0.40 | 2.78 | 2.82 | 2.80 |
| 0.60 | 3.42 | 3.44 | 3.43 |
| 0.80 | 3.94 | 3.98 | 3.96 |
| 1.00 | 4.42 | 4.44 | 4.43 |
Find the acceleration of the falling ball.
Step 1 - Work out the mean speed at each height
- At 0.20 m, (1.96 + 2.00 + 1.98) ÷ 3 = 1.98 m/s.
- At 0.40 m, (2.78 + 2.82 + 2.80) ÷ 3 = 2.80 m/s.
- At 0.60 m, (3.42 + 3.44 + 3.43) ÷ 3 = 3.43 m/s.
- At 0.80 m, (3.94 + 3.98 + 3.96) ÷ 3 = 3.96 m/s.
- At 1.00 m, (4.42 + 4.44 + 4.43) ÷ 3 = 4.43 m/s.
Step 2 - Square each mean speed
- At 0.20 m, 1.982 = 3.92 m2/s2.
- At 0.40 m, 2.802 = 7.84 m2/s2.
- At 0.60 m, 3.432 = 11.76 m2/s2.
- At 0.80 m, 3.962 = 15.68 m2/s2.
- At 1.00 m, 4.432 = 19.62 m2/s2.
Step 3 - Plot the graph
Plot (mean speed)2 on the y-axis against height on the x-axis. The points lie on a straight line through the origin, as predicts.
Step 4 - Find the gradient
Step 5 - Calculate the acceleration from the gradient
Comparing with , the gradient is .
This matches the accepted gravitational field strength of 9.8 N/kg.
Step 6 - Explain the effect of air resistance
Air resistance acts upwards, so the resultant force on the ball is slightly less than its weight, and its acceleration slightly less than . Over a drop of a metre this is too small to show in these results.
Common exam questions
Question 1
Why is the square of the speed plotted rather than the speed itself?
Because v2 = 2as, v2 is directly proportional to the height and gives a straight line through the origin. Speed against height would give a curve, with no single gradient to find the acceleration from.
Question 2
Why is the diameter of the ball entered into the datalogger?
The light gate times how long the beam is broken. The diameter tells the datalogger how far the ball moved in that time, so it can work out a speed.
Question 3
Why is the ball released rather than thrown downwards?
So it starts from rest, which makes u = 0. That's what shortens v2 = u2 + 2as to v2 = 2as.
Question 4
A tennis ball and a heavier ball of the same size are dropped together. Which reaches the ground first, and why?
The heavier one, though only just. Both have the same air resistance at a given speed, but it's a smaller fraction of the heavier ball's weight, so it accelerates slightly more.
Question 5
A student's graph of v2 against height is a straight line, but it doesn't pass through the origin. Suggest why.
The heights were probably measured from the wrong zero, such as the floor rather than the beam of the light gate, so every height is out by the same amount. This systematic error moves the whole line without changing its gradient.