A uniform horizontal rod of length l falls vertically from height h on two identical blocks placed symmertrically below the rod as shown in figure the coefficients of restitution are `e _(1)` and `e_(2)` the maximum height through which the centre of mass of the rod will rise after after bouncing off the blocks is

Options

A. `h/((e_(1)+e_(2))`
B. `((e_(1)+e_(2))^(2)h)/4`
C. `((e_(1)+e_(2))^(2)h)/2`
D. `(4h)/((e_(1)^(2)+e_(2)^(2))`

Answer

B

Explanation

`v_(cm)=(e_(1)+e_(2))/2sqrt(2gh)`
`h_(max)=(v_(cm))^(2)/(2g) = ((e_(1)+e_(2))^(2)h)/4`