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 4R. The angle of projection \theta is then given by : [1-Feb-2024 Shift 1]

Correct Answer is : sin−1[2gT2π2R]12

\sin ^{-1} [\frac{2 g T^2}{\pi^2 R}] ^{\frac{1}{2}}

\sin ^{-1} [\frac{\pi^2 R}{2 {\gT}^2}] ^{\frac{1}{2}}

\cos ^{-1} [ {2 g {\T}^2}/{\pi^2 {\R}}] ^{\frac{1}{2}}

\cos ^{-1} [ {\pi {\R}}/{2 {\gT}^2}] ^{\frac{1}{2}}

Circular Motion

Section: Physics

Question : 18 of 72

Marks: +1, -0

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 4R. The angle of projection θ is then given by :

[1-Feb-2024 Shift 1]

sin‌−1[‌
2gT2
π2R
]‌
1
2
sin‌−1[‌
π2R
2gT2
]‌
1
2
cos−1[‌
2gT2
π2R
]‌
1
2
cos−1[‌
πR
2gT2
]‌
1
2

Validate

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