A projectile is projected with velocity of \(25\, m / s\) 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 \(g =10 m / s ^{2}\) ]

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Correct Answer

(D) \(\cot ^{-1}\left(\frac{ R }{20 t ^{2}}\right)\)

Step-by-step Solution

Detailed explanation

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

\(t =\frac{ V \sin \theta}{ g } \Rightarrow V =\frac{ gt }{\sin \theta}\)

\(\Rightarrow R =\frac{ g ^{2} t ^{2}}{\sin ^{2} \theta} \cdot \frac{2 \sin \theta \cos \theta}{ g }\)

\(\tan \theta=\frac{2 gt ^{2}}{ R }=\frac{20 t ^{2}}{ R }\)

\(\cot \theta=\frac{ R }{20 t ^{2}}\)

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