Kinematics

JEE Main level, single correct, medium

216 of 308

Question 216 of 308

A particle moving in a circle of radius RR with uniform speed takes time TT to complete one revolution. If this particle is projected with the same speed at an angle θ\theta to the horizontal, the maximum height attained by it is equal to 4R4R. The angle of projection θ\theta is then given by :

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Answer: (A) sin⁡−1[2gT2π2R]12{\sin }^{-1}{\left[\dfrac{2g{T}^{2}}{{\pi }^{2}R}\right]}^{\dfrac{1}{2}}

Time period of circular motion is given by,
2πRv=T⇒v=2πRT\dfrac{2\pi R}{v}=T \Rightarrow v=\dfrac{2\pi R}{T}
Now, maximum height H=v2sin⁡2θ2gH=\dfrac{{v}^{2}{\sin }^{2}\theta }{2g}
⇒4R=4π2R2T2sin⁡2θ2g\Rightarrow 4R=\dfrac{4{\pi }^{2}{R}^{2}}{{T}^{2}}\dfrac{{\sin }^{2}\theta }{2g}
⇒sin⁡θ=2gT2π2R\Rightarrow \sin \theta =\sqrt{\dfrac{2g{T}^{2}}{{\pi }^{2}R}}
⇒θ=sin⁡−1(2gT2π2R)12\Rightarrow \theta ={\sin }^{-1}{\left(\dfrac{2g{T}^{2}}{{\pi }^{2}R}\right)}^{\dfrac{1}{2}}

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