Question

A ball which is at a rest is dropped from a height h meter. If it bounces off the floor it's speed is 80% of what it was just before touching the ground. The ball then rise to nearly at a height

Answer 1

Hey.

The ball at rest is dropped from height h m

so, u = 0 m/s , s = h m , a = g

As,
${v}^{2} = {u}^{2} + 2as \\ \\ so \: \: {v}^{2} = 0 + 2 \times g \times h \\ \\ so \: \: v \: = \sqrt{2gh}$
Now after rebound , the velocity reduced to 80% of the initial

so,
$new \: u \: = 80\% \: of \: \sqrt{2gh} \\ \\ or \: u = 0.80 \times \sqrt{2gh}$
and final velocity at new height s will be
v = 0 m/s

so,
${v}^{2} - {u}^{2} \: = 2( - g)h \\ as \: g \: is \: retarding \: here \\ \\ or \: \: 0 - {(0.80 \sqrt{2gh})}^{2} = - 2gs \\ \\ or \: \: s \: = \frac{({0.80 \sqrt{2gh})}^{2}}{2g} \: \\ \\ or \: \: s \: = \frac{0.64 \times 2gh}{2g} \\ \\ so \: \: s \: = 0.64h \\ \\ so \: \: s \: = 64\% \: of \: h$

So, the ball will rise nearly to a height 64 % of the initial h meters.

Thanks.

Answer 2

Answer:

0.64h

Explanation:

e=v of separation/ v of approach

so 0.8v/v

=0.8

h=e^2n.ho

 =(0.8)^2.h

=0.64h