A wedge of height 'h' is released from rest with a light particle `P` placed on it as shown. The wedge slides down an incline which makes an angle theta with the horizontal. All the surface are smooth, `P` will reach the surface of the incline in time:

Options

A. `sqrt((2h)/(gsin^(2)theta))`
B. `sqrt((2h)/(gsinthetacostheta))`
C. `sqrt((2h)/(g tantheta))`
D. `sqrt((2h)/(g cos^(2)theta))`

Answer

A

Explanation

Acceleration of wedge `=gsintheta`
FBD of particle w.r.t. wedge
`x((h)/(tantheta))`
Acceleration of particle along dispalcement `=gsinthetacostheta`
`t=sqrt((2s)/(a))=sqrt((2//htantheta)/(gsinthetacostheta))=sqrt((2h)/(gsin^(2)theta))` .