Question

A block slides with constant velocity on a plane inclined at an angle $\theta$. The same block is pushed up the plane with an initial velocity $v_0$. The distance covered by the block before coming to rest is :-

• A. $\frac{v_0^2}{2g\sin \theta }$
• B. $\frac{v_0^2}{4g\sin \theta }$
• C. $\frac{v_0^2{\sin }^2\theta }{2g}$
• D. $\frac{v_0^2{\sin }^2\theta }{4g}$

Answer 1

The correct option is A $\frac{v_0^2}{2g\sin \theta }$

Using third eq $v^2=u^2+2as$

where v is final velocity which is Zero, v_{0} is the initial velocity upward along the inclined plane.

$a=gsin\theta$ downward along the inclined plane.

$0=v_0^2-2gsin\theta s$

$s=\frac{v_0^2}{2gsin\theta }$