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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 $g =10 m / s ^{2}$ ]
$\frac{1}{2} \sin ^{-1}\left(\frac{5 t^{2}}{4 R}\right)$
$\frac{1}{2} \sin ^{-1}\left(\frac{4 R }{5 t ^{2}}\right)$
$\tan ^{-1}\left(\frac{4 t ^{2}}{5 R }\right)$
$\cot ^{-1}\left(\frac{ R }{20 t ^{2}}\right)$
Solution
$R =\frac{ V ^{2}(2 \sin \theta \cos \theta)}{ g }$
$t =\frac{ V \sin \theta}{ g } \Rightarrow V =\frac{ gt }{\sin \theta}$
$\Rightarrow R =\frac{ g ^{2} t ^{2}}{\sin ^{2} \theta} \cdot \frac{2 \sin \theta \cos \theta}{ g }$
$\tan \theta=\frac{2 gt ^{2}}{ R }=\frac{20 t ^{2}}{ R }$
$\cot \theta=\frac{ R }{20 t ^{2}}$
