A projectile is projected with velocity of 25   m   s - 1 at an angle θ with the horizontal.…

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,

$t=\frac{T}{2}=\frac{u\sin \theta }{g}$.

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

Now,

$$\begin{gathered} \frac{R}{t^2}=\frac{2u^2\sin \theta \cos \theta }{g}\times \frac{g^2}{u^2{\sin }^2\theta }=2g\cot \theta \\ \Rightarrow \theta ={\cot }^{-1}\left(\frac{R}{20t^2}\right) \end{gathered}$$