In the previous problem , the maximum height attained by the ball and the distance traveled by the block , after collision will be

Options

A. `(2 L)/(9) ,(4 L)/(9)`
B. `(4 L)/(3) ,(L)/(3)`
C. `(4L)/(9) ,(L)/(9)`
D. `(L)/(9),(2L)/(9)`

Answer

C

Explanation

For the ball
`0 = v_(1)^(2) - 2 gh`
` 0 = (4)/(9) xx 2gL - 2 gh`
`h = (4 L)/(9)`
For the block
`0 = v_(2)^(2) - 2 mu gd = (1)/(9) xx 2 g L - 2 xx 1 xx gd`
`d = (L)/(9)`