For a given velocity, a projectile has the same range R for two angles of projection, if $t_{1} a n d t_{2}$ are the times of flight in the two cases then

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Solution

The correct option is B

As we know for complementary angles will have the same range for a projectile with a given velocity.
Therefore,
$t_{1} = \frac{2 u \times s i n θ_{1}}{g}$
$t_{2} = \frac{2 u \times s i n θ_{2}}{g}$
$s i n θ_{1} + s i n θ_{2} = 90^{o}$
$t_{1} t_{2} = \frac{2 R}{g} ∴ t_{1} t_{2} \propto R$
Where,
$R = \frac{(u)^{2} \times s i n 2 θ}{g}$

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