Question

A rock is kicked horizontally at a speed of 16 m/s from the edge of a cliff. The rock strikes the ground 65 m from the foot of the cliff of height H as suggested in the figure, which is not drawn to scale. Neglect air resistance.

Answer 1

Answer:

First, let's solve the horizontal problem.

The horizontal velocity is 16m/s, and there is no forces in this axis, so the velocity is constant:

v(t) = 16m/s.

Now, we know that the rock hits the ground 65m from the foot of the cliff.

We can use the relation:

time = distance/velocity = 65m/16m/s = 4.06 seconds.

This is the time that the rock needs to hit the ground.

Now let's analyze the vertical problem.

The only force acting on the rock is the gravitational force, then we can write the acceleration as:

a(t) = -g

where g = 9.8m/s^2

For the velocity, we should integrate over time and get:

v(t) = -g*t + v0

where v0 is the initial velocity in this axis, but we do not have any, so v0 = 0.

v(t) = -g*t

Now, to get the position we should integrate again over time, and get:

p(t) = (-g/2)*t^2 + p0

where p0 is the initial vertical position, in this case is p0 = H.

p(t) = -(9.8/2 m/s^2)*t^2 + H

And we know that at t= 4.06 seconds, the rock hits the ground, then:

p(4.06s) = 0m =  -(9.8/2 m/s^2)*(4.06s)^2 + H

                       = -80.77m + H

H = 80.77m

The height of the cliff is 80.77 meters.