A particle of mass `m` is projected up from the bottom of an inclined plane with initial velocity `v_(0)` at angle `45^(@)` with an inclined plane of inclination `30^(@)` as shown in figure. At the same time a small block of same mass `m` is released from rest at a height `h`, the particle hits the block at some point on the inclined plane. Neglect friction at inclined.
Q. The value of height h is

Options

A. `(v_(0)^(2))/g`
B. `(sqrt(3)v_(0)^(2))/g`
C. `(2v_(0)^(2))/(sqrt(3)g)`
D. `(v_(0)^(2))/(sqrt(3)g)`

Answer

D

Explanation

`T=(2V_(0)sin45^(@))/(2gcos30)=(2sqrt(2)V_(0))/(sqrt(3)g)`
Since `x_(1)+x_(2)2=2h`
`implies ((V_(0))/(sqrt(2))(2sqrt(2)v_(0))/(sqrt(3)g)-1/2 gsin 30^(@)xx(8v_(0)^(2))/(3g^(2)))+(1/2 g/2 (8v_(0)^(2))/(3g))=2h`
`h=(v_(0)^(2))/(sqrt(3)g)`