A ball is projected from ground with a speed 15 ,ms ^{-1} at an angle θ with horizontal so that its

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3-2.Motion in Plane

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A ball is projected from the ground with a speed $15 \,ms ^{-1}$ at an angle $\theta$ with horizontal so that its range and maximum height are equal, then $tan\,\theta$ will be equal to

A

$\frac{1}{4}$

B

$\frac{1}{2}$

C

$2$

D

$4$

(JEE MAIN-2022)

Solution

$R = H$

$\frac{2 v _{ x } \times v _{ y }}{ g }=\frac{ v _{ y }^{2}}{2 g }$

$v _{ x }=\frac{ v _{ y }}{4} ; u \cos \theta=\frac{ u \sin \theta}{4}$

$\tan \theta=4$

Standard 11

Physics

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