A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it is equal to 4 R . The angle of projection $\theta$ is then given by :

(A) $\sin ^{-1}\left[\frac{2 g T^{2}}{\pi^{2} R}\right]^{\frac{1}{2}}$

(B) $\sin ^{-1}\left[\frac{\pi^{2} R}{2 g T^{2}}\right]^{\frac{1}{2}}$

(C) $\cos ^{-1}\left[\frac{2 \mathrm{gT}^{2}}{\pi^{2} \mathrm{R}}\right]^{\frac{1}{2}}$

(D) $\cos ^{-1}\left[\frac{\pi \mathrm{R}}{2 \mathrm{gT}^{2}}\right]^{\frac{1}{2}}$

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Kinematics

A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle

θ

to the horizontal, the maximum height attained by it is equal to 4 R . The angle of projection

θ

is then given by :

(A) $sin^{- 1} [\frac{2 g T ^{2}}{π ^{2} R}]^{\frac{1}{2}}$ ✓ Correct
(B) $sin^{- 1} [\frac{π ^{2} R}{2 g T ^{2}}]^{\frac{1}{2}}$
(C) $cos^{- 1} [\frac{2 gT ^{2}}{π ^{2} R}]^{\frac{1}{2}}$
(D) $cos^{- 1} [\frac{π R}{2 gT ^{2}}]^{\frac{1}{2}}$

Answer: (A)

Solution

Time period

T = \frac{2 π R}{v}

gives the particle's speed

v = \frac{2 π R}{T}

. Projected at angle

θ

with this speed, the maximum height reached is

H = \frac{v ^{2} sin ^{2} θ}{2 g} = 4 R ⟹ \frac{4 π ^{2} R ^{2}}{T ^{2}} \cdot \frac{sin ^{2} θ}{2 g} = 4 R ⟹ sin^{2} θ = \frac{2 g T ^{2}}{π ^{2} R}

θ = sin^{- 1} [\frac{2 g T ^{2}}{π ^{2} R}]^{1/2}

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option

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JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Kinematics

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