Question 230 of 308

The initial speed of a projectile fired from ground is $u$. At the highest point during its motion, the speed of projectile is $\dfrac{\sqrt{3}}{2}u$. The time of flight of the projectile is:

A$\dfrac{u}{2g}$B$\dfrac{u}{g}$C$\dfrac{2u}{g}$D$\dfrac{\sqrt{3}u}{g}$

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Answer: (B) $\dfrac{u}{g}$

At maximum height, only horizontal velocity of the particle remains. Let the angle of project be $\theta$. The horizontal velocity of the particle will remain same, therefore
$u\cos \theta =\dfrac{\sqrt{3}u}{2} \Rightarrow \theta =30^{\circ}$
Now, time of flight of projectile is
$T=\dfrac{2u\sin \theta }{g}=\dfrac{u}{g}$