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Question Number 55190 by Tawa1 last updated on 19/Feb/19

	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Feb/19

	$0 = u s i n α - {g T}_{1}$ $0 = u s i n β - {g T}_{2}$ $\frac{u s i n α}{u s i n β} = \frac{{g T}_{1}}{{g T}_{2}}$ $\frac{T_{1}}{T_{2}} = \frac{s i n α}{s i n β} . . . . (1)$ $x = (u c o s α) t_{1}$ $y = (u s i n α) t_{1} - \frac{1}{2} {g t}_{1}^{2}$ $c o m m o n p o i n t s o (x, y) i d e n t i c a l f o r t w o p a r t i c l e$ $x = (u c o s β) t_{2}$ $y = (u s i n β) t_{2} - \frac{1}{2} {g t}_{2}^{2}$ $(u c o s α) t_{1} = (u c o s β) t_{2}$ $\frac{c o s α}{c o s β} = \frac{t_{2}}{t_{1}}$ $\frac{t_{1}}{c o s β} = \frac{t_{2}}{c o s α} = k (s a y)$ $(u s i n α) t_{1} - \frac{1}{2} {g t}_{1}^{2} = (u s i n β) t_{2} - \frac{1}{2} {g t}_{2}^{2}$ $u s i n α \times k c o s β - u s i n β \times k c o s α = \frac{g}{2} (k c o s β + k c o s α) (k c o s β - k c o s α)$ $u k s i n (α - β) = \frac{g}{2} \times k^{2} \times [2 c o s (\frac{α + β}{2}) c o s (\frac{α - β}{2}) \times 2 s i n (\frac{α + β}{2}) s i n (\frac{α - β}{2})$ $e l i m i n a t i n g s i n (α - β) f r o m b o t h s i d e . .$ $u = \frac{g k}{2} \times s i n (α + β)$ $k = \frac{2 u}{g s i n (α + β)} . . . . (2)$ $t_{1} = \frac{2 u c o s β}{g s i n (α + β)} t_{2} = \frac{2 u c o s α}{g s i n (α + β)}$ $L H S$ $T_{1} t_{1} + T_{2} t_{2}$ $= \frac{u s i n α}{g} \times \frac{2 u c o s β}{g s i n (α + β)} + \frac{u s i n β}{g} \times \frac{2 u c o s α}{g s i n (α + β)}$ $\frac{2 u^{2}}{g^{2} s i n (α + β)} \times [s i n α c o s β + c o s α s i n β]$ $\frac{2 u^{2}}{g^{2}} \times \frac{s i n (α + β)}{s i n (α + β)}$ $= \frac{2 u^{2}}{g^{2}} . . . p r o v e d . . . .$
	Commented by Tawa1 last updated on 19/Feb/19

	$God bless you sir . I really appreciate your effort$
	Commented by Tawa1 last updated on 23/Feb/19

	$Sir, i {don}^{'} t understand how you eliminate \sin (α - β) sir .$
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