A projectile is projected with velocity of $$25$$ 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 $$g = 10$$ m s$$^{-2}$$]

We are given a projectile with initial velocity $$u = 25$$ m/s at angle $$\theta$$ with the horizontal, and after time $$t$$ seconds its inclination with horizontal becomes zero (i.e., it reaches the highest point). We need to find $$\theta$$ in terms of $$t$$ and range $$R$$.

Find time to reach highest point: at the highest point, the vertical component of velocity is zero:

$$ u\sin\theta - gt = 0 $$

$$ 25\sin\theta = 10t $$

$$ \sin\theta = \frac{2t}{5} \quad \cdots (1) $$

Write the range formula: $$ R = \frac{u^2 \sin 2\theta}{g} = \frac{625 \times 2\sin\theta\cos\theta}{10} = 125\sin\theta\cos\theta $$

Express $$\cos\theta$$ from the range equation: substituting $$\sin\theta = \frac{2t}{5}$$ from equation (1):

$$ R = 125 \times \frac{2t}{5} \times \cos\theta = 50t\cos\theta $$

$$ \cos\theta = \frac{R}{50t} \quad \cdots (2) $$

Find $$\tan\theta$$: dividing equation (1) by equation (2):

$$ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{2t/5}{R/(50t)} = \frac{2t}{5} \times \frac{50t}{R} = \frac{20t^2}{R} $$

Express $$\theta$$: $$ \theta = \tan^{-1}\left(\frac{20t^2}{R}\right) = \cot^{-1}\left(\frac{R}{20t^2}\right) $$

Therefore, the correct answer is Option D.