A projectile can have the same range R for two angles of projection. If $T_{1}$ and $T_{2}$ be the times of flights in the two cases, then the product of the two times of flights is directly proportional to:

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$\frac{1}{R^{2}}$

$\frac{1}{R}$

$R$

$R^{2}$

answer is C.

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Detailed Solution

$r a n g e o f p r o j e c t i l e i s s a m e f o r c o m p l e m e n t a r y a n g l e s i . e . f o r θ a n d (90^{0} - θ) . time of flight for θangle of projection = T_{1} = \frac{2 u \sin θ}{g} time of flight for(90- θ)angle of projection = T_{2} = \frac{2 u \sin (90^{0} - θ)}{g} = \frac{2 u \cos θ}{g}$

$range of a projectile= R = \frac{u^{2} \sin 2 θ}{g} p r o d u c t o f t i m e o f f l i g h t s i s, T_{1} T_{2} = \frac{2 u \sin θ}{g} \times \frac{2 u \cos θ}{g} = \frac{2 u^{2} (2 \sin θ \cos θ)}{g^{2}} = \frac{2 u^{2} (\sin 2 θ)}{g^{2}} = \frac{2 R}{g}$