Question

The speed at the maximum height of a projectile is $\frac{\sqrt{3}}{2}$ times of the initial speed is $u^{\prime }$ of projection. Its range on the horizontal plane

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

Answer 1

Answer: (A)

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

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

$\theta =$ angle of projection

At the maximum height, vertical component is vanished and the horizontal component is $ucos\theta$

$ucos\theta =\frac{\sqrt{3}u}{2}$

$cos\theta =\frac{\sqrt{3}}{2}$

$\theta =co{s}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$\theta ={30}^0$

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

$R=\frac{u^{2}sin{60}^0}{g}$

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

The correct option is A.