For a given velocity, a projectile has the same range R for two angles of projection. If t₁ and t₂ are the times of flight in the two cases then

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Solution

The correct option is B
For same range angles of projection should be $θ$ and 90 - $θ$
So, time of flights $t_{1}$ = $\frac{2 u s i n θ}{g}$ and
$t_{2}$ = $\frac{2 u s i n (90 - θ)}{g}$ = $\frac{2 u c o s θ}{g}$
By multiplying = $t_{1} t_{2}$ = $\frac{4 u^{2} s i n θ c o s θ}{g^{2}}$
$t_{1} t_{2}$ = $\frac{2}{g}$ $\frac{(u^{2} s i n 2 θ)}{g}$ = $\frac{2 R}{g}$ $\Rightarrow$ $t_{1} t_{2}$ $α$ R

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