Horizontal range of a projectile is 4√ 3 times its maximum height. Its angle of projection will be

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3-2.Motion in Plane

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The horizontal range of a projectile is $4\sqrt 3 $ times its maximum height. Its angle of projection will be ......... $^o$

A

${45}$

B

${60}$

C

${90}$

D

${30}$

Solution

(d) $R = 4H\cot \theta ,$

If $R =4\sqrt 3 H$ then $\cot \theta = \sqrt 3 $

$⇒$ $\theta = 30^\circ $

Standard 11

Physics

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