Two bodies are projected from the same point with equal speeds in such directions that they both strike the same point on a plane whose inclination is$\beta$. If $\propto$ be the angle of projection of the first body with the horizontal the ratio of their times of flight is

Question

Dr. Abubacker · Accepted Answer

(b) Let $\alpha^{''}$ be the angle of projection of the second body
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$R = \frac{u^{2}}{gcos^{2}\beta}\left\lbrack \sin(2\alpha - \beta) - sin\beta ightbrack$
Range of both the body is same. Therefore
$\sin(2\alpha - \beta) = sin\left( 2\alpha^{''} - \beta ight)$
$or\ 2\alpha^{''} - \beta = \pi - (2\alpha - \beta)$
$\alpha^{''} = \frac{\pi}{2} - (\alpha - \beta)$
$Now,\ \ \ \ T = \frac{2u\sin(\alpha - \beta)}{gcos\beta}\ and\ T^{''} = \frac{2u\sin\left( \alpha^{''} - \beta ight)}{gcos\beta}$
$	herefore\ \frac{T}{T^{''}} = \frac{\sin(\alpha - \beta)}{\sin\left( \alpha^{''} - \beta ight)} = \frac{\sin(\alpha - \beta)}{\sin\left\{ \frac{\pi}{2} - (\alpha - \beta) - \beta ight\}}$
$\frac{\sin(\alpha - \beta)}{\sin\left( \frac{\pi}{2} - \alpha ight)} = \frac{\sin(\alpha - \beta)}{cos\alpha}$

Q. Two bodies are projected from the same point with equal speeds in such directions that they both strike the same point on a plane whose inclination is$\beta$. If $\propto$ be the angle of projection of the first body with the horizontal the ratio of their times of flight is

Explanation

Reviewed by Dr. Abubacker

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