Question

The speed at the maximum height of a projectile is $\frac{\sqrt{3}}{2}$ times of its initial speed '$u$' 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

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

speed at the maximum height is only going to be horizontal component which is $ucos\theta$

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

$\theta ={30}^o$

therefore range $R=\frac{u^{2}sin2\theta }{g}=\frac{u^{2}sin{60}^o}{g}=\frac{\sqrt{3}{u}^2}{2g}$