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3-2.Motion in Plane
hard
A particle moving in a circle of radius $R$ with uniform speed takes time $\mathrm{T}$ 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$\sin ^{-1}\left[\frac{\pi^2 R}{2 \mathrm{gT}^2}\right]^{\frac{1}{2}}$
C$\cos ^{-1}\left[\frac{2 \mathrm{gT}^2}{\pi^2 \mathrm{R}}\right]^{\frac{1}{2}}$
D$\cos ^{-1}\left[\frac{\pi R}{2 g T^2}\right]^{\frac{1}{2}}$
(JEE MAIN-2024)
Solution
$\frac{2 \pi R}{T}=V$
$\text { Maximum height } H=\frac{v^2 \sin ^2 \theta}{2 g}$
$4 R=\frac{4 \pi^2 R^2}{T^2 2 g} \sin ^2 \theta$
$\sin \theta=\sqrt{\frac{2 g T^2}{\pi^2 R}}$
$\theta=\sin ^{-1}\left(\frac{2 g T^2}{\pi^2 R}\right)^{\frac{1}{2}}$
$\text { Maximum height } H=\frac{v^2 \sin ^2 \theta}{2 g}$
$4 R=\frac{4 \pi^2 R^2}{T^2 2 g} \sin ^2 \theta$
$\sin \theta=\sqrt{\frac{2 g T^2}{\pi^2 R}}$
$\theta=\sin ^{-1}\left(\frac{2 g T^2}{\pi^2 R}\right)^{\frac{1}{2}}$
Standard 11
Physics
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