Question

A ball is dropped from a height of $20$m on a floor for which $e=1/2$. The height attained by the ball after the second collision.

• A. 1.25 m
• B. 2.5 m
• C. 5 m
• D. 10 m

Answer 1

The correct option is A 1.25 m

Initially it reaches ground with a velocity of $\sqrt{2gh}=20m/s$

Let the velocity of separation after first collision be x

And the velocity of collision after second collision be y,

Then as per equation of e,

$\frac{x}{20}=\frac{1}{2}$ and $\frac{y}{x}=\frac{1}{2}$

Solving we get y=5m/s

So height attained by ball after second collision = $\frac{y^2}{2g}=1.25m$