Question

The velocity at the maximum height of a projectile is $\frac{\sqrt{3}}{2}$ times its initial velocity of projection $\left(u\right)$. Its range on the horizontal plane is

• A. $\frac{\sqrt{3}{u}^2}{2g}$
• B. $\frac{3u^2}{2g}$
• C. $\frac{3u^2}{g}$
• D. $\frac{u^2}{2g}$

Answer 1

The correct option is A $\frac{\sqrt{3}{u}^2}{2g}$

As the vertical component of velocity becomes $0$ at maximum height $\left(H\right)$. Therefore, velocity at maximum height comprises only of horizontal component of velocity.
$\Rightarrow v_H=u\cos \theta$
According to question,
$u\cos \theta =\frac{\sqrt{3}}{2}u$
$\Rightarrow \theta ={30}^{\circ }$
$R=\frac{u^2\sin 2\theta }{g}=\frac{u^2\sin {60}^{\circ }}{g}$
$=\frac{\sqrt{3}{u}^2}{2g}$
Hence, the correct answer is option (a)