Question

The speed of projectile at its highest point is observed to be half of its speed of projection $u$. Its range on horizontal plane is-

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

Answer 1

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

Lets consider, $u=$ initial velocity of the projectile

$\theta =$ angle of projection

At maximum height,

$velocity=\frac{u}{2}$

$ucos\theta =\frac{u}{2}$

$cos\theta =0.5$

$\theta =co{s}^{-1}\left(0.5\right)$

$\theta ={60}^0$

Horizontal range, $R=\frac{u^{2}sin2\theta }{g}$

$R=\frac{u^{2}sin{120}^0}{g}$

$R=\frac{\sqrt{3}}{2}\frac{u^2}{g}$

The correct option is B.