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 $\mathrm{T}$ 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 $4 \mathrm{R}$. 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) $\sin ^{-1}\left[\frac{\pi^{2} \mathrm{R}}{2 \mathrm{gT}^{2}}\right]^{\frac{1}{2}}$
(3) $\cos ^{-1}\left[\frac{2 \mathrm{gT}^{2}}{\pi^{2} \mathrm{R}}\right]^{\frac{1}{2}}$
(4) $\cos ^{-1}\left[\frac{\pi \mathrm{R}}{2 \mathrm{gT}^{2}}\right]^{\frac{1}{2}}$
Show Answer
Answer: (1)
Solution:
$\frac{2 \pi \mathrm{R}}{\mathrm{T}}=\mathrm{V}$
Maximum height $\mathrm{H}=\frac{\mathrm{v}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}$
$4 \mathrm{R}=\frac{4 \pi^{2} \mathrm{R}^{2}}{\mathrm{~T}^{2} 2 \mathrm{~g}} \sin ^{2} \theta$
$\sin \theta=\sqrt{\frac{2 \mathrm{gT}^{2}}{\pi^{2} \mathrm{R}}}$
$\theta=\sin ^{-1}\left(\frac{2 \mathrm{gT}^{2}}{\pi^{2} \mathrm{R}}\right)^{\frac{1}{2}}$
