Question

A ball is thrown vertically upwards from the edge of the cliff and hits the ground at the base of the cliff with a speed of 30 m/s four seconds later. How high was the cliff?

Answer 1

To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.

$v = v_0 -gt$

Here,

v = Final velocity

$v_0$ = Initial velocity

g = Acceleration due to gravity

t = Time

At t = 4s, v = -30m/s (Downward)

Therefore the initial velocity will be

$-30 = v_0 -9.8(4)$

$v_0 = 9.2m/s$

Now the position can be calculated as,

$y = h +v_0t -\frac{1}{2}gt^2$

When it has the ground, y=0 and the time is t=4s,

$0 = h+(9.2)(4)-\frac{1}{2} (9.8)(4)^2$

$h = 41.6m$

Therefore the cliff was initially to 41.6m from the ground