Question

A ball is projected from an inclined plane having inclination $\theta$, normal to the plane with speed $u$. It strikes the vertical wall horizontally and again it stikes the inclined plane. If all the collisions are elastic, find the time taken by the ball to hit the inclined plane.

• A. $T=\frac{2u\sin \theta }{g}$
• B. $T=\frac{2u\cos \theta }{g}$
• C. $T=\frac{2u\tan \theta }{g}$
• D. $T=\frac{u\sin \theta }{g}$

Answer 1

The correct option is B $T=\frac{2u\cos \theta }{g}$


The ball strikes the wall horizontally which means vertical component of speed $=0$
Hence, by $v=u+at$
$0=u\cos \theta -gt$
or $t=\frac{u\cos \theta }{g}$
Here, $t=$ time taken by the ball to hit the vertical wall.

Given, collision is elastic.
Hence, velocity of seperation = velocity of approach
$0-v=(u\sin \theta -0)$
$\Rightarrow v=-u\sin \theta$
(-ve sign means ball will change its direction by ${180}^{\circ })$

As the direction of ball changed by ${180}^{\circ }$ after hitting the wall, the ball will hit the inclined plane at the point of projection after following back the same path.
Hence, total time taken by the ball to hit the inclined surface again is
$T=2\times t$
$T=\frac{2u\cos \theta }{g}$