Free Fall Class Exercises
Physics
1. Wile E Coyote steps off a VERY tall cliff and sits still. After he realizes that he is not standing on the ground anymore, he free falls toward the ground below.
Time
Speed
1 s
9.8 m/s
2 s
19.6 m/s
3 s
29.4 m/s
4 s
39.2 m/s
5 s
49.0 m/s
Distance Fallen 4.9 m 19.6 m 44.1 m 78.4 m 122.5 m
Free fall happens when gravity is the only significant force acting on an object. All object in free fall accelerate downward at 9.8 m/s/s.
Most falling objects have non-neglible amounts of air resistance which means they are not in free fall. If air resistance (or any other force) is significant, then the object will either accelerate at a lower rate or move at a constant speed. For our purposes, we always assume that there is no air affecting a moving object unless the situation specifically mentions the air.
After he notices that he’s not standing on the ground, Wile E. Coyote free falls , so we know that his accelerating downward at 9.8 m/s/s. Because he is standing still before he starts falling, we know that his v i
= 0 m/s.
At 1 sec: v f
= v i
+ at v f
= 0 m/s + (9.8 m/s/s)(1 s) v i
= 9.8 m/s
At 2 sec: v f
= v i
+ at v v i f
= 0 m/s + (9.8 m/s/s)(2 s)
= 19.6 m/s d = v i
·t + Ошибка!
at
2 d = (0 m/s)·(1 s) + Ошибка!
(9.8 m/s/s)(1 s) 2 d = 4.9 m d = v f
·t – Ошибка!
at
2 d = (0 m/s)·(2 s) + Ошибка!
(9.8 m/s/s)(2 s)
2 d = 19.6 m
2. steps off a 15 m high cliff overlooking the lake and free falls until hitting the water. a) How long will it take them to hit the ground? d = v i
·t + Ошибка!
at
2
15 m = (0 m/s)·t + Ошибка!
(9.8 m/s/s)∙t 2
15 m = 4.9 m/s/s∙t 2
3.06 s
2
= t
2
1.75 s = t b) How fast will they be moving when they hit the water? v f
= v i
+ at v f
= 0 m/s + (9.8 m/s/s)(1.75 s) v i
= 17.15 m/s
3. is crossing a 70 m deep ravine on a rope bridge, when the rope snaps and drops them toward the river below.

a) How long do they have before they hit the water? b) How fast will they be traveling when they hit the water?
[t = 3.78 s] [v f
= 37.04 m/s]
4. A crow who wants to eat a snail will drop it from a high place to crack the snail’s shell and expose the tasty flesh. In order to crack the snail’s shell, it must be going 15 m/s when it hits the ground. Assume that the snail free falls. a) How high does the crow have to be when it drops the snail for the shell to crack? b) How long does the snail’s fall from that height last?
[d = 11.48 m]
5. a) One second after they leave the trampoline:
[t = 1.53 s]
jumps on a trampoline. They leave the trampoline traveling upward at 12 m/s. i) How high above the trampoline are they?
This situation is different from the previous problems because the object (or person, in this case) launched going upward. This doesn’t change the nature of free fall – the object is still accelerating is downward at 9.8 m/s/s.
The only adjustment we have to make is to use signs indicate which direction things are going. I choose up be the positive direction – that makes the initial upward velocity positive and the downward acceleration negative. This makes sense because we know that an object which is launched upward will eventually stop and start coming back down – it has to slowing down (negative acceleration) to come to a stop. to to be d = v i
·t + Ошибка!
at
2 d = (+12 m/s)∙(1 s) + Ошибка!
(–9.8 m/s/s)∙(1s)
2 d = 12 m +–4.9 m ii) How fast are they moving? v f
= v i
+ at = +12 m/s + (–9.8 m/s/s)(1 s) = +2.2 m/s = 2.2 m/s going up b) Two seconds after they leave the trampoline, i) How high above the trampoline are they? ii) How fast are they moving?
[d = 4.4 m] [v f
= –7.6 m/s]

c) When they reach the highest point in their jump, how high above the trampoline are they?
You know you’re at the highest point in a flight when you stop moving – that’s when you turn around and start coming back down. Mathematically, you know you’re at the top when your speed is zero. In this case, we end up at the top, so v f
= 0 m/s. v f
2 = v i
2 + 2a·d
(0 m/s) 2 = (+12 m/s) 2 + 2(–9.8 m/s/s)·d
0 m
2
/s
2
= 144 m
2
/s
2
–19.6 m/s/s·d
–144 m 2 /s 2 = –19.6 m/s/s·d
7.35 m = d
6. The best human jumpers in the world have a vertical leap of about 1.2 m a) How fast would they have to leave the ground to get that high at the top of their jump? b) How long will they be in the air (up and down)?
(Hint: where do they land compared to where they took off?)
[v i
= 4.85 m/s] [t = 4.85 m/s]
7. gets up on the 3 m high diving board to do a dive. They bounce on the board and get launched upward at 7 m/s. a) How long will it take before they hit the water below?
The distance d = v i
·t + Ошибка!
at
2
–3 m = (+7 m/s)∙t + Ошибка!
( –9.8 m/s/s) t
2
0 = ( –4.9 m/s/s)∙ t 2 + (7 m/s)∙t +3 m
But this is simply a quadratic equation. Remember that any equation that looks like: 0 = a∙t 2 + b∙t + c has the solution: t =
Ошибка!
In our specific equation, a = –4.9 m/s/s; b = 7 m/s; c = 3 m. t =
Ошибка!
=
Ошибка!
=
Ошибка!
=
Ошибка!
That gives us two answers: t =
Ошибка!
=
Ошибка!
= –0.35 s t =
Ошибка!
=
Ошибка!
= 1.77 s
We have to think about the answer and choose the best one. In this case it is easy, because negative time would mean that the diver hit the water before they jumped off the diving board. That’s ridiculous, so we choose the positive answer: t = 1.77 s b) How fast will they be going when they get there? v f
= v i
+ at = +7 m/s + ( –9.8 m/s/s )∙(1.77 s) = – 10.35 s ≈ –10.38 m/s
8. A gymnast dismounts from the high bar of the uneven bars, when her body is 2.3 m above the ground. She leaves the bars moving upward at 5 m/s.

a) How fast will she be moving when she hits the ground?
v f
2
= v i
2
+ 2a·d v f
2
= (+5 m/s)
2
+ 2( –9.8 m/s/s) ·(
–2.3 m) v f
2
= 70.08 m
2
/s
2
We take the square root to solve for v f
, but we have to remember that there are two square roots of any number – a positive one and a negative one. In the past, we ignored this because everything was going in the positive direction, but we have to think about this now that things are moving up and down. v f
= +8.37 m/s or v f
= –8.37 m/s
We need to choose the right one. In this case, we know that the diver is moving downward when they hit the water. Since up is the positive direction, we need a negative number to indicate that they are moving downward, so we choose our second answer: v f
= –8.37 m/s b) How much time will she have in the air to do her twisting double flip? v f
= v i
+ at
–8.37 m/s = +5 m/s + ( –9.8 m/s/s )∙t
–13.37 m/s = ( –9.8 m/s/s )∙t
1.36 s = t
Notice that in problem 7 and problem 8, we solved the same questions – how long did the flight last and how fast was the object going when it landed – using two different methods. Both of those methods are perfectly valid and you may use whichever method you are more comfortable with when faced with this type of problem.
9. During a snowball fight, your friend hides in a tree, 3 m off the ground. You sneak up under the tree and throw a snowball straight up at them with all the strength you can muster. The snowball leaves your hand going 25 m/s. a) How much time will your friend have to dodge the snowball (if they notice it at the moment you throw it)? b) How fast will the snowball be going when it hits your friend?
[t = 0.123 s] [v f
= 23.79 m/s]