Question

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

Difficult

Solution

See fig. (3)
Let α be the angle of projection of the second body. The range of both the bodies will be the same. Therefore,
$\begin{array}{l} R = \frac{u^{2}}{{gcos}^{2} β} [\sin (2 α - β) - \sin β] \\ or \sin (2 α - β) = \sin (2 α^{'} - β) \\ or \begin{array}{l} (2 α^{'} - β) = π - (2 α - β) \\ Now, T = \frac{2 usin (α - β)}{gcos β} and T^{'} = \frac{2 usin (α^{'} - β)}{gcos β} \\ ∴ \frac{T}{T^{'}} = \frac{\sin (α - β)}{\sin (α^{'} - β)} = \frac{\sin (α - β)}{\sin [\frac{π}{2} - (α - β) - β]} \\ = \frac{\sin (α - β)}{\sin (\frac{π}{2} - α)} = \frac{\sin (α - β)}{\cos α} \end{array} \end{array}$

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