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

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Q.
A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were 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 [2021]

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

2 $θ = \cos^{- 1} {(\frac{g T^{2}}{π^{2} R})}^{1 / 2}$

3 $θ = \cos^{- 1} {(\frac{π^{2} R}{g T^{2}})}^{1 / 2}$

4 $θ = \sin^{- 1} {(\frac{π^{2} R}{g T^{2}})}^{1 / 2}$

Ans.
(1)

We know, $T = \frac{2 π R}{u} \Rightarrow u = \frac{2 π R}{T}$

Here : $H_{max} = 4 R$

$H_{max} = \frac{u^{2} \sin^{2} θ}{2 g}$ ...(i)

Putting the value of $u$ and $H_{max}$ in equation (i), we get

$4 R = {(\frac{2 π R}{T})}^{2} \times \frac{\sin^{2} θ}{2 g} \Rightarrow θ = \sin^{- 1} {(\frac{2 g T^{2}}{π^{2} R})}^{1 / 2}$

Solution

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

2 $θ = \cos^{- 1} {(\frac{g T^{2}}{π^{2} R})}^{1 / 2}$

3 $θ = \cos^{- 1} {(\frac{π^{2} R}{g T^{2}})}^{1 / 2}$

4 $θ = \sin^{- 1} {(\frac{π^{2} R}{g T^{2}})}^{1 / 2}$

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Solution

Q. A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were 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 [2021]

1 θ=sin-1(2gT2π2R)1/2

2 θ=cos-1(gT2π2R)1/2

3 θ=cos-1(π2RgT2)1/2

4 θ=sin-1(π2RgT2)1/2

Ans. (1) We know, T=2πRu⇒u=2πRT Here : Hmax=4R Hmax=u2sin2θ2g ...(i) Putting the value of u and Hmax in equation (i), we get 4R=(2πRT)2×sin2θ2g ⇒ θ=sin-1(2gT2π2R)1/2

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

2 $θ = \cos^{- 1} {(\frac{g T^{2}}{π^{2} R})}^{1 / 2}$

3 $θ = \cos^{- 1} {(\frac{π^{2} R}{g T^{2}})}^{1 / 2}$

4 $θ = \sin^{- 1} {(\frac{π^{2} R}{g T^{2}})}^{1 / 2}$