Question

A stone is dropped off a cliff and hit the ground with a speed of 120 ft/s . a) what is the equation for acceleration a(t), velocity v(t), and displacement/ position s(t)? b) what is the height of the cliff ?

Answer 1

The key to tackling this problem is remembering that if the stone is dropped and not thrown, it implies that the initial velocity of the stone was zero. So the problem has given you the final and initial velocities.

so for part a) it becomes much simpler. If we solve: \[V (final) = V (initial) +at\] for a \[(V (final) - V (initial))/t = a\]

Answer 2

First, write down everything you know (Note: i = initial and f = final)

Vi = 0 ft/s
Vf = 120 ft/s
a = -32 ft/s^2 (acceleration due to gravity)
Xi = 0 ft/s (I picked this as our starting point)

From there, first need to know the time it took for the stone to fall which you can use the following kinematic equation to find:

Vf = Vi + a * t -> this equates to 120 ft/s = 0 ft/s + (-32 ft/s^2) * t

Solve for t = to get 3.75 seconds

Now that you have the fall time, you can solve for the height of the cliff using the kinematic equation:
Xf -Xi = Vi +1/2gt^2 -> this equates to: Xf - 0ft = 0 ft/s + (-32ft/s) * 3.75s

Solve for Xf to get -120ft which indicates a 120 foot drop

I think this should get you going to providing the equations the problem asks for.

Good luck!