1. Interpret the condition

The inclination of the projectile with the horizontal becomes zero when its velocity is purely horizontal.

That happens at the highest point, where the vertical component of velocity becomes zero.

So, $u_y - gt = 0$ where initial speed $u=25\,\text{m/s}$.

Thus, $25\sin\theta - 10t = 0$ $\sin\theta = \frac{2t}{5}$

2. Use the formula for range

Horizontal range of a projectile is $R = \frac{u^2\sin 2\theta}{g}$ Substituting $u=25$ and $g=10$, $R = \frac{25^2\sin 2\theta}{10}$ $R = \frac{625}{10}\sin 2\theta$ $R = \frac{125}{2}\sin 2\theta$

So, $\sin 2\theta = \frac{2R}{125}$

But using $\sin 2\theta = 2\sin\theta\cos\theta$ with $\sin\theta = \frac{2t}{5}$ we get $\sin 2\theta = 2\cdot \frac{2t}{5}\cdot \cos\theta = \frac{4t}{5}\cos\theta$ Hence, $\frac{4t}{5}\cos\theta = \frac{2R}{125}$ $\cos\theta = \frac{R}{50t}$

3. Find $\tan\theta$

Now, $\tan\theta = \frac{\sin\theta}{\cos\theta}$ $\tan\theta = \frac{\frac{2t}{5}}{\frac{R}{50t}}$ $\tan\theta = \frac{2t}{5}\cdot \frac{50t}{R}$ $\tan\theta = \frac{20t^2}{R}$

Therefore, $\theta = \tan^{-1}\left(\frac{20t^2}{R}\right)$

Using the identity $\tan^{-1}(x)=\cot^{-1}\left(\frac{1}{x}\right)$ we get $\theta = \cot^{-1}\left(\frac{R}{20t^2}\right)$

4. Match with options

This matches Option D: $\cot ^{ - 1}\left( \frac{R}{20t^2} \right)$

5. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.