Question

A particle is projected from a point at the foot of a fixed plane inclined at angle ${45}^{\circ }$ to the horizontal, in the vertical plane containing the line of greatest slope through the point. If $\theta (>{45}^{\circ }$) is the inclination to the horizontal of the initial direction of projection, for what values of $\tan \theta$ will the particle strike the plane horizontally?

Answer 1

Ignore the inclined plane for this problem.

Angle of projection with the horizontal is $\theta$.
As the final velocity is parallel to the horizontal plane, the projectile is at its maximum height from the horizontal plane.

At maximum height, angle of elevation of the projectile = ${45}^{\circ }$
$\Rightarrow \tan {45}^{\circ }=\frac{H}{\frac{R}{2}}\Rightarrow R=2H$

By using the relation $4H=R\tan \theta$,
$4H=2H\tan \theta$
$\therefore \tan \theta =2$