A projectile is projected with velocity of 25m s ^ - 1 at an angle...

From the information given in the question, after time $t$ the inclination of the projectile becomes zero i.e. the projectile is at the highest point. Therefore,

$t = \dfrac{T}{2} = \dfrac{usin\theta}{g}$.

The rang of the projectile $R = \dfrac{2{u}^{2}sin\theta cos\theta}{g}$.

Now,

$\dfrac{R}{{t}^{2}} = \dfrac{2{u}^{2}sin\theta cos\theta}{g} \times \dfrac{{g}^{2}}{{u}^{2}\sin^{2}\theta} = 2gcot\theta ⇒ \theta = \cot^{- 1}\left(\dfrac{R}{20{t}^{2}}\right)$