Question

A ball is projected with velocity $u$ at right angle to the slope which is inclined at an angle $\alpha$ with the horizontal. The distance ${}^{\prime }{x}^{\prime }$ along the inclined plane that it will travel before again striking the slope is-

• A.
  $\frac{2u^2}{g}\cos \alpha$
• B.
  $\frac{2u^2}{g}\frac{\tan \alpha }{\sin \alpha }$
• C.
  $\frac{2u^2}{g}\frac{\tan \alpha }{\cos \alpha }$
• D.
  $\frac{2u^2}{g}\tan \alpha$

Answer 1

The correct option is C

$\frac{2u^2}{g}\frac{\tan \alpha }{\cos \alpha }$

Step 1: Draw diagram for the given situation.

Step 2:Find time taken by the ball again strike the slope.

Equation for motion perpendicular to the incline,
$y=0$
$y=u.t-\frac{1}{2}g\cos \alpha .t^2$
$\Rightarrow t=\frac{2u}{g\cos \alpha }$

Step 3: Find the distance along the incline when ball strike the slope
As we know, initial velocity along the incline,
$u_x=0$
And acceleration along the incline,
$a_x=g\sin \alpha$
Apply equation of motion along the incline,
$x=ut+\frac{1}{2}g{t}^2$
$x=0+\frac{1}{2}\times \left(g\sin \alpha \right){\left(\frac{2u}{g\cos \alpha }\right)}^2$
$x=\frac{2u^2}{g}\frac{\tan \alpha }{\cos \alpha }$

Final answer: (C)