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 $\alpha$ 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

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