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 $β$ . If $α$ be the angle of projection of the first body with the horizontal the ratio of their times of flight is:

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$\frac{c o s (α - β)}{s i n α}$

$\frac{\cos α}{\sin (α - β)}$

$\frac{\sin (α - β)}{\cos α}$

$\frac{\cos α}{\sin α}$

answer is C.

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Detailed Solution

. $L e t α' b e t h e a n g l e o f p r o j e c t i o n o f t h e s e c o n d b o d y r a n g e = R = \frac{u^{2}}{g \cos β} [\sin (2 α - β) - \sin β] \sin (2 α - β) = \sin (2 α^{'} - β) o r 2 α^{'} - β = π - (2 α - β) \Rightarrow α^{'} = \frac{π}{2} - (α - β)$

$t i m e o f f l i g h t = T = \frac{2 u \sin (α - β)}{g \cos β} a n d T^{1} = \frac{2 u \sin (α^{1} - β)}{g \cos β} d i v i d e t h e t w o e q u a t i o n s \frac{T}{T^{1}} = \frac{\sin (α - β)}{\sin (α^{1} - β)} \frac{T}{T^{1}} = \frac{\sin (α - β)}{\sin \{\frac{π}{2} - (α - β) - β\}} \frac{T}{T^{1}} = \frac{\sin (α - β)}{\sin (\frac{π}{2} - α)} \frac{T}{T^{1}} = \frac{\sin (α - β)}{\cos α}$