A projectile has a range R and time of flight T. If the range is doubled by increasing the speed of projection, without changing the angle of projection, the time of flight will become

$\frac{T}{\sqrt{2}}$

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$\sqrt{2} T$

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$\frac{T}{2}$

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$2 T$

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Solution

The correct option is B
$\sqrt{2} T$

We know that, $R = \frac{v_{0}^{2} s i n 2 θ_{0}}{g}$
When the range is doubled without changing the angle of projection, the initial velocity becomes $\sqrt{2}$ times.
Now, $T = \frac{2 v_{0} s i n θ_{0}}{g} = 2 \times \sqrt{\frac{R g}{s i n 2 θ_{0}}} \times \frac{s i n θ_{0}}{g}$
As $T \propto \sqrt{R}$ , the time of flight will become $\sqrt{2}$ times.
Hence, the correct choice is (b).

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