A ball is let fall from a height `h_(0)`. There are `n` collisions with the earth. If the velocity of rebound after `n` collisions is `v_(n)` and the ball rises to a height `h_(n)` then coefficient of restitution `e` is given by

Options

A. `e^(n)=sqrt((h_(n))/(h_(0)))`
B. `e^(n)=sqrt((h_(0))/(h_(n)))`
C. `"ne"=sqrt((h_(0))/(h_(0)))`
D. `sqrt("ne")=sqrt((h_(n))/(h_(0)))`

Answer

A

Explanation

In this problem the velocity of the earth before and after the collision may be assumed zero. Hence coefficient of restitution will be
`e^(n)=(v_(1))/(v_(0))xx(v_(2))/(v_(1))xx(v_(3))/(v_(2))xx…xx(v_(n))/(v_(n)-1)`
where `v_(n)` is the velocity after `nth` rebounding and `v_(0)` is the velocity with which the ball strikes the earth for the first time.
Hence
`e^(n)=(v_(n))/(v_(0))=(sqrt(2gh_(n)))/(sqrt(2gh_(0)))`
where `h_(n)` is the height two which the ball rises after `nth` rebounding. Hence
`e^(n)=(v_(n))/(v_(0))=(sqrt(h_(n)))/(sqrt(h_(0)))`