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

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Q.
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 [2024]

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

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

3 $\cos^{- 1} {[\frac{2 g T^{2}}{π^{2} R}]}^{\frac{1}{2}}$

4 $\cos^{- 1} {[\frac{π R}{2 g T^{2}}]}^{\frac{1}{2}}$

Ans.
(1)

$\frac{2 π R}{T} = v$

Maximum height $H = \frac{v^{2} \sin^{2} θ}{2 g}$

$4 R = \frac{4 π^{2} R^{2}}{T^{2} 2 g} \sin^{2} θ$

$\sin θ = \sqrt{\frac{2 g T^{2}}{π^{2} R}} \Rightarrow θ = \sin^{- 1} {(\frac{2 g T^{2}}{π^{2} R})}^{\frac{1}{2}}$

Solution

Q.
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 [2024]

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

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

3 $\cos^{- 1} {[\frac{2 g T^{2}}{π^{2} R}]}^{\frac{1}{2}}$

4 $\cos^{- 1} {[\frac{π R}{2 g T^{2}}]}^{\frac{1}{2}}$

Ans.
(1)

$\frac{2 π R}{T} = v$

Maximum height $H = \frac{v^{2} \sin^{2} θ}{2 g}$

$4 R = \frac{4 π^{2} R^{2}}{T^{2} 2 g} \sin^{2} θ$

$\sin θ = \sqrt{\frac{2 g T^{2}}{π^{2} R}} \Rightarrow θ = \sin^{- 1} {(\frac{2 g T^{2}}{π^{2} R})}^{\frac{1}{2}}$

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Solution

Q. 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 [2024]

1 sin-1[2gT2π2R]12

2 sin-1[π2R2gT2]12

3 cos-1[2gT2π2R]12

4 cos-1[πR2gT2]12

Ans. (1) 2πRT=v Maximum height H=v2sin2θ2g 4R=4π2R2T22gsin2θ sinθ=2gT2π2R⇒θ=sin-1(2gT2π2R)12

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Q.
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 [2024]

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

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

3 $\cos^{- 1} {[\frac{2 g T^{2}}{π^{2} R}]}^{\frac{1}{2}}$

4 $\cos^{- 1} {[\frac{π R}{2 g T^{2}}]}^{\frac{1}{2}}$

Ans.
(1)

$\frac{2 π R}{T} = v$

Maximum height $H = \frac{v^{2} \sin^{2} θ}{2 g}$

$4 R = \frac{4 π^{2} R^{2}}{T^{2} 2 g} \sin^{2} θ$

$\sin θ = \sqrt{\frac{2 g T^{2}}{π^{2} R}} \Rightarrow θ = \sin^{- 1} {(\frac{2 g T^{2}}{π^{2} R})}^{\frac{1}{2}}$