Practice Problems: Free Fall Solutions

1. (easy) A small ball is released from a window at t = 0. Assuming free-fall conditions, how far does it travel in 2.8 seconds? If the ball had more mass would it fall a greater distance? 
Assume y_o = 0
y - y_o = v_ot + ½gt²
y - 0 = 0 +-4.9(2.8)²
y = -38.4 m
Free-fall acceleration is independent of mass, so the distance fallen by a more massive object would be the same.

2. A rock is dropped from a garage roof from rest. The roof is 6.0 m from the ground.
a. (easy) Determine how long it takes the rock to hit the ground.
y - y_o = v_ot + ½at²
-6 - 0 = 0 + ½(-9.8)t²
t = 1.1 s
b. (easy) Determine the velocity of the rock as it hits the ground.
v = v_o + at
v = 0 + (-9.8)(1.1)
v = -10.8 m/s
c. (moderate) A second rock is projected straight upward from ground level at the moment the first rock was released. This second rock had an initial upward velocity of +6.0 m/s. How long will this second rock take to reach maximum height?
v = v_o + at
At max height v = 0
0 = 6 + (-9.8)t
t = 0.61 s
d. (hard) At what time after release will the two rocks have the same height?
y₁ - y_o,1 = v_o,1t + ½at²   AND  y₂ - y_0,2 = v_o,2 t + ½at²
When y₁ = y₂ ...  y_o,1 + v_o,1t + ½at² =  y_0,2 + v_o,2 t + ½at²
0 + 0 + (-4.9)t² = -6 + 6t + (-4.9)t²
6 = 6t
t = 1s

3. (moderate) A projectile is shot upward at 189 m/s from a height of 20 m off the ground. How long will it take for the projectile to be at ground level?
y - y_o = v_ot + ½at²
0 - 20 = 189t - 4.9t²
4.9t² - 189t - 20 = 0
Use the quadratic equation:
t = {38.7s, -0.1 s}
The root that makes physical sense is t = 38.7 s

4. (moderate) A hot-air balloon is hovering over a large public park. The operator of the balloon then makes the balloon begin to rise at a constant rate of 0.8 m/s. At some height from the ground the operator drops a rock from the basket. The rock takes 10.3 seconds to hit a target on the ground. How high (above the ground) was the rock when it was released and what was its maximum height?
To find initial height:
y - y_o = v_ot + ½at²
0 - y_o = 0.8(10.3) - 4.9(10.3)²
y_o = 511.60 m
To find max height: v = 0
v² = v_o² + 2gΔy
0 = (0.8)² - 19.6(y_max - 511.60)
y_max = 511.63 m  (the rock moved 3 cm upward before falling downward)

5. Design an experiment that would allow you to prove the following claim: "All objects accelerate toward the Earth at the same rate." Your answer should include a procedure (including a list of apparatuses needed to perform the experiment) that a fellow student could follow to collect the necessary data. Finally, state the types of calculations that could be made with that data to prove the claim.
Of course, there are many different procedures that can be used, and yours may be different than the one listed here. But your procedure should allow the student investigator to come to the same conclusion.
Apparatus needed: A measuring device (such as a tape measure), and a stopwatch
Procedure:
Step 1: Take two objects (preferrably two different massed dense metal spheres) to the second floor window of a home.
Step 2: Drop object 1 (from rest) while measuring the time it takes to fall from a specific height to the ground. Measure the height with a measuring tape.
Step 3: Drop the second object from the same height as the first, again measureing the time to hit the ground.
Step 4: Repeat steps 2 and 3 several times to average the time readings for each object.
Step 5. Repeat steps 2 through 4 for several different heights, perhaps use a third floor window if possible.
Analysis:
For each trial use kinematic equations to calculate the acceleration of the object in that trial (ex: Δx = v_ot + ½at²).
Conclusion: Each trial will show the same acceleration for each object. Any discrepencies would be small and within reason based on differences in the performance of each trial. For example, the air resistance may vary slightly based on the wind conditions, but the overall effect of air resistance can be reasonably neglected over short distances.

6. (moderate) A model rocket enthusiast launches a rocket with a motion sensor in the launchpad. Assume y = 0 at the launchpad, that positive is up, and that the fuel mass is very small compared to the rocket body mass. Create qualitative y-t, v-t, and a-t graphs for an experiment that starts at lift off and ends when the rocket hits the Earth on the way back down. Assume free-fall after the rocket fuel is used up.


7. (moderate) Determine the distance between two steel spheres (after 1.4 s) dropped from a tower if the second sphere was dropped 0.5 seconds after the first. Assume free-fall and that the spheres are dropped from rest. 
(Use y_o = 0 and + is up)
Sphere 1 after 1.4 s:
y₁ = y_o + v_ot + ½gt²
y₁ = 0 + 0 - 4.9(1.4)² 
y₁ = -9.6 m
Sphere 2 after 0.9 s:
y₂ = y_o + v_ot + ½gt²
y₂ = 0 + 0 - 4.9(0.9)² 
y₁ = -4.0 m
The spheres will be 5.6 m apart at that time.

8. (easy) Watch the video below that shows a hammer and a feather being dropped by an astronaut standing on the moon. If the acceleration due to gravity on the moon is 1/6 the acceleration due to gravity on the Earth, how long did the feather take to fall 1.0 m? Additionally, how would the results of this experiment differ if done on Earth?

9. (OPTIONAL hard) If a ball is tossed up (free-fall conditions) with an initial speed of 2.0 m/s does it spend more time in the top 0.1 m of the toss or the bottom 0.1 m of the toss?
(Use y_o = 0 and + is up)
First find the max height (where v = 0):
v² = v_o² + 2gΔy
0 = 2² - 19.6Δy
Δy = max height = 0.2 m
Thus, the top motion is from y = 0.1 m (on the way up) to y = 0.1 (on the way down):
We need to find the speed at y = 0.1:
v² = v_o² + 2gΔy
v² = 2² - 19.6(0.1)
v = 1.4 m/s
(by symmetry the velocity at y = 0.1 on the way down is opposite)
Now find time for the top motion:
v = v_o + gt
-1.4 = 1.4 - 9.8t
t = 0.29s
For the bottom motion on the way up:
y - y_o = ½(v + v_o)t
0.1 - 0 = ½(1.4 + 2)t
t = 0.06 s
By symmetry, for the bottom motion on the way down, t = 0.06 s.
Thus, the top of the motion lasts 0.29 s while the bottom of the motion lasts 0.12 s. This helps to explain why the volleyball player with a big vertical leap can appear to hover in the air.
The visualization shown below may help your understanding.


Click on a link below to see Scenario Solutions:

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