Question

A ball is dropped from rest at a height of 15m.if it loses 30% of kinetic energy on striking the ground, what is the height to which it bounces?

Answer 1

Answer:

The height from which the ball falls gives it a potential energy which converts to kinetic energy as the ball falls down. When the ball finally hits the ground, all the potential energy has converted to kinetic energy. If we assume mass of ball as M kg, then

Mgh = (final kinetic energy)

120M = (final K.E.)

Ball loses 25% of (final K.E.), so initial energy with which it bounces back = 75% of (final K.E) = (3/4) × 120M = 90M.

So to find the height to which the ball will bounce back we equate the initial energy (that is 90M) with the final energy (that is potential energy at the maximum height to which the ball has bounced).

90M = MgH

H= 9 metres.

So that's your answer.

There was a kinetic energy loss because you know, when the ball will bounce it'll create some sound and heat and vibrations on the floor, so the 25% of kinetic energy is lost there.

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Answer 2

The height to which it bounces is 10.5 m.

Explanation:

Use conservation of energy.

When the ball  hits the ground, all the potential energy has converted to kinetic energy as

Mgh = final kinetic enegy (K)

Here, h is the height and g is the acceleration due to gravity.

$Mg\times15m=K$

As 30% of the kinetic energy is lost on striking, the retained kinetic energy after the impact is 70%.

so, 70% of final kintic enegy (K) $=\frac{7}{10}(Mg\times15m)$

It is converted into PE at the maximum height .

At the maximum height the body only poses potential energy when it bounces back.

Therefore,

$\frac{7}{10}(Mg15) =MgH$

$H=\frac{7}{10}\times15m$

H = 10.5 m.

Thus, the height to which it bounces is 10.5 m.

#Learn More: conservation of energy.

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