NEET · Physics · STD 11 - 3.2 motion in plane
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 ' \(\theta\) ' to the horizontal, the maximum height attained by it equals \(4 \mathrm{R}\). The angle of projection, \(\theta\), is then given by
- A \(\theta=\cos ^{-1}\left(\frac{g T^{2}}{\pi^{2} R}\right)^{1 / 2}\)
- B \(\theta=\cos ^{-1}\left(\frac{\pi^{2} R}{g T^{2}}\right)^{1 / 2}\)
- C \(\theta=\sin ^{-1}\left(\frac{\pi^{2} R}{g T^{2}}\right)^{1 / 2}\)
- D \(\theta=\sin ^{-1}\left(\frac{2 \mathrm{~g} T^{2}}{\pi^{2} R}\right)^{1 / 2}\)
Answer & Solution
Correct Answer
(D) \(\theta=\sin ^{-1}\left(\frac{2 \mathrm{~g} T^{2}}{\pi^{2} R}\right)^{1 / 2}\)
Step-by-step Solution
Detailed explanation
\(T=\frac{2 \pi R}{V}\) \(V=\frac{2 \pi R}{T}\)
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