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A projectile is projected with velocity of 25 m/s at an angle $θ$ with the horizontal. After t seconds its inclination with horizontal become zero. If R represents horizontal range of the projectile, the value of $θ$ will be [Use $g = 10 m / s^{2}$ ]

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

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

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

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

answer is D.

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Detailed Solution

$R = \frac{V^{2} (2 \sin θ \cos θ)}{g}$

$t = \frac{Vsin θ}{g} \Rightarrow V = \frac{gt}{\sin θ}$

$\Rightarrow R = \frac{g^{2} t^{2}}{\sin^{2} θ} \cdot \frac{2 \sin θ \cos θ}{g}$

$\tan θ = \frac{2 g t^{2}}{R} = \frac{20 t^{2}}{R}$

$\cot θ = \frac{R}{20 t^{2}}$

$θ = \cot^{- 1} (\frac{R}{20 t^{2}})$

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