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,
$R = \frac{v_{0}^{2} sin 2 θ_{0}}{g}$

When the range is doubled without changing the angle of projection, the initial velocity becomes $\sqrt{2}$ times.
Now,
$T = \frac{2 v_{0} sin θ_{0}}{g} = 2 \times \sqrt{\frac{R g}{sin 2 θ_{0}}} \times \frac{sin θ_{0}}{g}$

As $T \propto \sqrt{R}$ , the time of flight will become $\sqrt{2}$ times.

Hence, the correct choice is $(b)$ .

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