Question

A projectile is thrown at an angle theta Such that it is just able to cross a vertical wall at heighest point of journey the angle theta at which the projectile is thrown is given by?

Answer 1

Maximum height is given by formula

$$h=\frac {{v_0}^2 {\sinθ}^2}{2g} (1)$$

Using (1) we get:

$$θ=\sqrt{\arcsin{\frac {2gh}{{v_0}^2 }}}$$