Question

A ball is dropped from the roof of a 25-m-tall building. What is the velocity of the object when it touches the ground? Suppose the ball is a perfect golf ball and it bounces such that the ve locity as it leaves the ground has the same magnitude but the op posite direction as the velocity with which it reached the ground How high will the ball bounce? Now suppose instead that the ball bounces back to a height of 20 m. What was the velocity with which it left the ground?

Answer 1

Answer:

a)  $h=25m$

b)  $v=19.8m/sec$

Explanation:

From the question we are told that:

Height $h=25m$

Bounce Height $h'=20m$

Generally the Kinematic equation is mathematically given by

$V=\sqrt{2gh}\\\\V=\sqrt{2*9.81*25}$

$V=22.1m/sec$

Therefore Height

$h=\frac{V^2}{2g}\\\\h=\frac{22.1^2}{2*9.81}$

$h=25m$

b)

Generally the Kinematic equation is mathematically given by

$v^2=2ah$

$v^2=2*9.8*20$

$v=\sqrt{2*9.8*20}$

$v=19.8m/sec$