Question

A particle is projected with velocity such that it just crosses two walls of height h and seprated by h Find the angle of projection.

Answer 1

Let velocity of projection be $v$ and angle of projection is $\theta$.

Separation between the wall is $\theta$

Time taken to cross both wall of each height h is equal to difference of instant when it would be at a height h above the ground.

The kinematics of motion of particle in vertical direction is

$y=y_o+u_{y}t+\frac{1}{2}{a}_y{t}^2$

$h=0+usin\theta -\frac{1}{2}g{t}^2$

$sin\theta =h+\frac{1}{2}g{t}^2$

$\theta ={sin}^{-1}(h+\frac{1}{2}g{t}^2)$