Question

A ball is dropped from a height of 100 m on the ground. If the coefficient of restitution is 0.2, the height up to which the ball will go after it rebounds for the 2nd time:
A) 16 m
B) 1.6 cm
C) 1.6 m
D) 16 cm ​

Answer 1

Given :

Height of building = 100m

Coefficient of restitution = 0.2

To Find :

Height attained by ball after second collision.

Solution :

❒ Suppose a body is dropped from a height h and it strikes the ground with velocity v. After the inelastic collision let it rise to a height h'. If v' be the velocity with which the body rebounds, then

$\dag\:\underline{\boxed{\bf{\gray{e=\dfrac{v'}{v}=\bigg(\dfrac{2gh'}{2gh}\bigg)^{\frac{1}{2}}=\bigg(\dfrac{h'}{h}\bigg)^{\frac{1}{2}}}}}}$

If after n collisions with the ground, the height to which it rises be $\sf{h_n}$, then

• $\bf\:e^n=\bigg(\dfrac{h_n}{h}\bigg)^{\frac{1}{2}}$

By substituting the given values;

$:\implies\sf\:(0.2)^2=\bigg(\dfrac{h_n}{100}\bigg)^{\frac{1}{2}}$

$:\implies\sf\:(0.04)^2=\dfrac{h_n}{100}$

$:\implies\sf\:h_n=0.0016\times 100$

$:\implies\:\underline{\boxed{\bf{\blue{h_n=0.16m=16cm}}}}$

∴ (D) is the correct answer!