Q 11-03-083JEE MainJEE Main 2023 (31 Jan, Shift 1)Easy
The initial speed of a projectile fired from ground is $u$. At the highest point during its motion, the speed of projectile is $\dfrac{\sqrt3}{2}u$. The time of flight of the projectile is
Answer: (B) $\dfrac ug$
At the top only $u\cos\theta$ remains: $u\cos\theta=\dfrac{\sqrt3}{2}u\Rightarrow\theta=30^\circ$.
$$T=\frac{2u\sin30^\circ}{g}=\frac ug$$
Solution by Sreeraj P, M.Sc Physics