JEE Mains · Physics · STD 11 - 3.2 motion in plane
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 \(\cos ^{-1}\left[\frac{\pi R}{2 g T^2}\right]^{\frac{1}{2}}\)
Answer & Solution
Correct Answer
(A) \(\sin ^{-1}\left[\frac{2 g T^2}{\pi^2 R}\right]^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
\(\frac{2 \pi R}{T}=V\) \(\text { Maximum height } H=\frac{v^2 \sin ^2 \theta}{2 g}\) \(4 R=\frac{4 \pi^2 R^2}{T^2 2 g} \sin ^2 \theta\) \(\sin \theta=\sqrt{\frac{2 g T^2}{\pi^2 R}}\) \(\theta=\sin ^{-1}\left(\frac{2 g T^2}{\pi^2 R}\right)^{\frac{1}{2}}\)
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