If time of flight of a bullet over a horizontal range R is T , then angle of projection with horizon

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3-2.Motion in Plane

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If the time of flight of a bullet over a horizontal range $R$ is $T$, then the angle of projection with horizontal is ......

A

$\tan ^{-1}\left(\frac{g T^2}{2 R}\right)$

B

$\tan ^{-1}\left(\frac{2 R^2}{g T}\right)$

C

$\tan ^{-1}\left(\frac{2 R}{g^2 T}\right)$

D

$\tan ^{-1}\left(\frac{2 R}{g T}\right)$

Solution

(a)

$T=\frac{2 u \sin \theta}{g} \Rightarrow u=\frac{g T}{2 \sin \theta}$

$R=\frac{2 u^2 \sin \theta \cos \theta}{g}$

$R=\frac{2 u \sin \theta}{g} \times u \cos \theta$

$R=T \times u \cos \theta$

$R=T \times \frac{g T \cos \theta}{2 \sin \theta}$

$R=\frac{g T^2}{2} \frac{1}{\tan \theta}$

$\tan \theta=\frac{g T^2}{2 R}$

$\theta=\tan ^{-1}\left(\frac{g T^2}{2 R}\right)$

Standard 11

Physics

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