Question

A projectile is projected from the edge of an inclined plane which is at an inclination of ${45}^o$ from the ground, with the velocity 'u' at an angle ${}^{\prime }{\theta }^{\prime }$ which is greater than ${45}^o$ with the horizontal in a vertical plane. What will be the value of ${}^{\prime }\tan {\theta }^{\prime }$ if the projectile strikes the plane normally?

• A. $1$
• B. $2$
• C. $\sqrt{3}$
• D. $\infty$

Answer 1

The correct option is B $2$

Acceleration along the plane $=\frac{g}{\sqrt{2}}$

velocity component along the plane $=\mu \cos \theta$

Since particles strikes perpendicular , its velocity component along the plane must have become zero

using $v=u+at$

where $v=0,u=\mu \cos \theta ,a=\frac{-g}{\sqrt{2}}$ and $t=\frac{2\mu \sin \theta }{\frac{g}{\sqrt{2}}}$

$0=\mu \cos \theta -\frac{g}{\sqrt{2}}\times \frac{2\mu \sin \theta }{\frac{g}{\sqrt{2}}}\tan \theta =2$