A block slides down a rough inclined plane of slope angle `theta` with a constnat velocity. It is then projected up the same plane with an intial velocity v the distance travelled by the block up the plane coming to rest is .

Options

A. `(v_(0)^(2))/(2g sin theta)`
B. `(v_(0)^(2))/(4g sin theta)`
C. `(v_(0)^(2)sin^(2)theta)/(2g)`
D. `(v_(0)^(2)sin^(2)theta)/(4g)`

Answer

B

Explanation

`mg sin theta=mu mg cos theta`
when pushed up
deacc `=g sin theta+mug cos theta`
`therefore v^(2)-v_(0)^(2)=2(a)xxs`
`therefore s=(v_(0)^(2))/(2(2g sin theta))`