Question

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

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

Answer 1

The correct option is C $\frac{2u^2}{g}\frac{\tan \alpha }{\cos \alpha }$


We know that range of a projectile down an incline plane is given by
$R=\frac{u^2}{g{\cos }^2\alpha }\left[\sin \right(2\theta +\alpha )+\sin \alpha ]$ (where $\theta$ is the angle of projection with the horizontal and $\alpha$ is the angle of inclination of the plane)
Here $\theta ={90}^{\circ }-\alpha$ and angle of inclination = $\alpha$
$\Rightarrow x=\frac{u^2}{g{\cos }^2\alpha }\left[\sin \right(2(90-a)+\alpha )+\sin \alpha ]=\frac{2u^2\sin \alpha }{g{\cos }^{2}a}=\frac{2u^2}{g}\frac{\tan \alpha }{\cos \alpha }$