A projectile is projected with velocity of 25 m/s at an angle $θ$ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of $θ$ will be : [use g = 10 m/s²]

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$\cot^{- 1} (\frac{R}{20 t^{2}})$

$\tan^{- 1} (\frac{4 t^{2}}{5 R})$

$\frac{1}{2} \sin^{- 1} (\frac{5 t^{2}}{4 R})$

$\frac{1}{2} \sin^{- 1} (\frac{4 R}{5 t^{2}})$

answer is D.

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Detailed Solution

$\begin{aligned} R = \frac{V^{2} (2 \sin θ \cos θ)}{g} \\ t = \frac{Vsin θ}{g} \Rightarrow V = \frac{gt}{\sin θ} \\ \Rightarrow R = \frac{g^{2} t^{2}}{\sin^{2} θ} \cdot \frac{2 \sin θ \cos θ}{g} \\ \tan θ = \frac{2 {gt}^{2}}{R} = \frac{20 t^{2}}{R} \\ \cot θ = \frac{R}{20 t^{2}} \end{aligned}$