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 time of flights is proportional to

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

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R^{2}

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R

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\frac{1}{R}

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Solution

The correct option is D $R$

Two projectile have same range if their angle of projections are complementary. Range $R = \frac{u^{2} s i n 2 θ}{g}$

$t_{1} = \frac{2 u s i n θ}{g} t_{1} = \frac{2 u s i n (90 - θ)}{g} t_{1} t_{2} = \frac{4 u^{2} s i n θ c o s θ}{g^{2}} = \frac{2 u^{2} s i n 2 θ}{g^{2}} = \frac{2 R}{g}$

The product of the two time of flights is proportional to R.

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A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flights in the two cases, then the product of the two time of flights is proportional to

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