Calculate the angle of projection at which a projectile with velocity √2gh is projectile so that it just crosses a wall of height

Question

Calculate the angle of projection at which a projectile with velocity\(\sqrt{2gh}\) is projectile so that it just crosses a wall of height a.h situated at a distance of h from the point of projection:

(a) 15°

(b) 75°

(c) 60°

(d) 30°

Answer 1

Correct option is : (d) 30°

For vertical upward motion:

\(y = (u\sin\alpha)t - \frac{1}{2}gt^2\)

\(\therefore\ h = (u\ \sin\alpha)t - \frac{1}{2}gt^2\)

or gt² - 2u sin α.t + 2h = 0 .......(1)

Solution of this quadratic equation.

If the particle crosses the wall in time t₁ and t₂ then time of flight,

Thus, option (d) is correct.