A particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution.

Question

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 $	heta$ to the horizontal, the maximum height attained by it equals $4R$. The angle of projection, $	heta$ is then given by:

Accepted Answer

1.  **Analyze Circular Motion:** The particle moves in a circle of radius $R$ with uniform speed $v$ and time period $T$. The speed is distance divided by time:
    $$ v = \frac{2\pi R}{T} $$
2.  **Analyze Projectile Motion:** The particle is projected with this same speed $v$ at an angle $	heta$. The maximum height $H$ attained in projectile motion is given by:
    $$ H = \frac{v^2 \sin^2 	heta}{2g} $$
3.  **Apply Given Condition:** We are given that the maximum height equals $4R$. So, $H = 4R$. Substituting the expression for $H$:
    $$ 4R = \frac{v^2 \sin^2 	heta}{2g} $$
4.  **Substitute and Solve:** Substitute the expression for $v$ derived in step 1 into the equation:
    $$ 4R = \frac{\left(\frac{2\pi R}{T}ight)^2 \sin^2 	heta}{2g} $$
    $$ 4R = \frac{\frac{4\pi^2 R^2}{T^2} \sin^2 	heta}{2g} $$
    $$ 4R = \frac{2\pi^2 R^2 \sin^2 	heta}{g T^2} $$
    Rearranging to solve for $\sin^2 	heta$:
    $$ \sin^2 	heta = \frac{4R \cdot g T^2}{2\pi^2 R^2} $$
    $$ \sin^2 	heta = \frac{2g T^2}{\pi^2 R} $$
    Taking the square root:
    $$ \sin 	heta = \left(\frac{2g T^2}{\pi^2 R}ight)^{1/2} $$
    $$ 	heta = \sin^{-1}\left[\left(\frac{2g T^2}{\pi^2 R}ight)^{1/2}ight] $$

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