Physics Rock Question

Question

A rock is thrown horizontally from a cliff with a speed of 15 m/s. It takes 4 seconds for the stone to fall on the ground.

A) How far does it fall from the bottom of the cliff?

B) How high is the cliff? (use g=9.81 m/s^2)

Answer 1

x = 15t

y = h - gt²/2;

A) t = 4 s, x = 15m/s·4s = 60 m

B) At t = 4 s y = h - gt²/2 = 0; h = gt²/2 = 9.81m/s²·16s²/2 = 78.48 m

Answer 2

This is a kinematics problem. Like the car it deals with uniformly accelerated motion. Don't let the mention of gravity fool you into thinking you need to be concerned with forces - we are dealing only with speed, acceleration and time! In this case, gravity is providing the acceleration, the same way that the car provided the acceleration in the other problem.

Big 4 Kinematic equations (it really is helpful to memorize these)

1) d = v_i*t + (1/2)at²

2) v_f² = v_i² + 2ad

3) v_f = v_i + at

4) d = [(v_i+ v_f) / 2]*t

What do we know from the question itself?

V_i = 15 m/s thrown horizontally

t = 4s to fall to the ground.

And the question wants us to determine how far the rock falls both horizontally and vertically.

The twist with this question is that we are now dealing with motion in 2 dimensions, so let's think about how the rock is moving. We are told that the rock is thrown horizontally, in the positive x direction, if that helps you when drawing a picture. Vertically, the rock starts at rest. All motion is horizontal initially. The rock is then accelerated by gravity in the vertical direction. While this is the same rock we are talking about, we can look at these two motions independently.

So let's get to work.

A) How far does it fall from the bottom of the cliff?

This question is asking us: how far has the rock traveled in the horizontal direction in 4s. What kinematic equation 1 through 4 do you think we should use? Since there is no horizontal acceleration, and we are looking for a relationship between distance, velocity and time, the answer is 1

1) d = v_i*t + (1/2)at²

Given that we are trying to determine the distance d, let's plug in what we know:

d = (15 m/s) * (4s) + (1/2)*(0)*(4s)²

d = 60m

That is our answer for part 1.

B) How high is the cliff? (use g=9.81 m/s^2)

Firstly, remember that this motion in the vertical direction is independent from the horizontal motion we just looked. Vertically, the rock is starting from rest. This question is then asking us: How far will gravity move a rock in 4 seconds?

We can again use the first equation:

1) d = v_i*t + (1/2)at²

We are again solving for a distance, except this time the initial velocity is 0, and the acceleration is 9.81 m/s2. Let's plug in some numbers:

d = (0 m/s) * 4s + (1/2)(9.81m/s)(4s)²

d = 78.5m

Hope this helps!