Question

The velocity at the maximum height of a projectile is half of its initial velocity of projection u. Its range on the horizontal plane is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

If the velocity of projection is u then at the highest point body posses only $ucos\theta$

$ucos\theta =\frac{u}{2}\left(given\right)\therefore \theta ={60}^{\circ }NowR=\frac{u^{2}sin(2\times {60}^{\circ })}{g}=\frac{\sqrt{3}}{2}\frac{u^2}{g}$