A projectile is projected with velocity of $25\ \text{m s}^{-1}$ at an angle $\theta$ with the horizontal. After $t$ seconds its inclination with horizontal becomes zero. If $R$ represents horizontal range of the projectile, the value of $\theta$ will be: [use $g = 10\ \text{m s}^{-2}$]
Answer: (D) $\cot^{-1}\left(\dfrac{R}{20t^2}\right)$
The velocity is horizontal at the highest point, so $t = \dfrac{u\sin\theta}{g}$, giving $u\sin\theta = 10t$.
Range: $R = \dfrac{2(u\sin\theta)(u\cos\theta)}{g} = \dfrac{2(10t)(u\cos\theta)}{10} = 2t\,u\cos\theta$, so $u\cos\theta = \dfrac{R}{2t}$.
$$\tan\theta = \frac{u\sin\theta}{u\cos\theta} = \frac{10t\cdot 2t}{R} = \frac{20t^2}{R}\ \Rightarrow\ \theta = \cot^{-1}\left(\frac{R}{20t^2}\right)$$
Solution by Sreeraj P, M.Sc Physics