Question

A ball is dropped from height h on a horizontal floor . If it loses 60% of its energy in hitting the floor then height upto which it will rise after first rebounce is

Answer 1

the velocity of ball just before hitting the floor is sqrt(2gh) and its kinetic energy will be 1/2mu^2=mgh.

after hitting it loses 60% of its energy or we say it possess only 40% of its initial energy .

its energy after hitting the floor is 40/100(mgh)

after reaching its maximum height h' it will lose all its kinetic energy and will possess only potential energy which is mgh'.

therefore law of conservation of energy says that
mgh'=40/100(mgh)
h'=0.4h

Answer 2

Using Eq law of motion

V^2 - U^2 = 2gh

V= root (2gh) .....U = 0 (freely falling)

Now From conservation of energy (COE)

Mgh= 1/2 MV^2

Since after falling the energy decreases by 40% in hitting the floor so new KE = 40/100x 1/2MV^2

This new KE is equal to PE of the next rebound

So if we apply COE

40/100 x1/2 MV^2 = Mgh1

2/5 x 1/2 V^2 = 10h1......

.......... (M get cancelled)

Putting V value as root (2gh)

2/5x1/2x2x10xh = 10 h1

2/5h = h1...... Ans