Motion In Two Dimensions Question 1
Question 1 - 2024 (01 Feb Shift 1)
A particle moving in a circle of radius $R$ with uniform speed takes time $T$ 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 :
(1) $\sin ^{-1}\left[\frac{2 g T^{2}}{\pi^{2} R}\right]^{\frac{1}{2}}$
(2) $\sin ^{-1}\left[\frac{\pi^{2} R}{2 gT^{2}}\right]^{\frac{1}{2}}$
(3) $\cos ^{-1}\left[\frac{2 gT^{2}}{\pi^{2} R}\right]^{\frac{1}{2}}$
(4) $\cos ^{-1}\left[\frac{\pi R}{2 gT^{2}}\right]^{\frac{1}{2}}$
Show Answer
Answer: (1)
Solution:
$\frac{2 \pi R}{T}=V$
Maximum height $H=\frac{v^{2} \sin ^{2} \theta}{2 g}$
$4 R=\frac{4 \pi^{2} R^{2}}{T^{2} 2 g} \sin ^{2} \theta$
$\sin \theta=\sqrt{\frac{2 gT^{2}}{\pi^{2} R}}$
$\theta=\sin ^{-1}\left(\frac{2 gT^{2}}{\pi^{2} R}\right)^{\frac{1}{2}}$
