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

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Kinematics

A ball is projected from the ground with a speed

15 ms^{- 1}

at an angle

θ

with horizontal so that its range and maximum height are equal, then

tan θ

will be equal to

(A) $\frac{1}{4}$
(B) $\frac{1}{2}$
(C)2
(D)4✓ Correct

Answer: (D)

Solution

For projectile motion, range

R = \frac{u ^{2} s i n 2 θ}{g}

and maximum height

H = \frac{u ^{2} s i n ^{2} θ}{2 g}

.

Given

R = H

\frac{u ^{2} sin 2 θ}{g} = \frac{u ^{2} sin ^{2} θ}{2 g}

2 sin θ cos θ = \frac{sin ^{2} θ}{2}

4 cos θ = sin θ

tan θ = 4

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option

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