Question

The velocity at the maximum height of a projectile is half its initial velocity of projection. Find its range on the horizontal plane. If $u=20{\text{ms}}^{-1}$. (Take $g=10{\text{m/s}}^2$)

• A. $20\sqrt{3}\text{m/s}$
• B. $10\sqrt{3}\text{m/s}$
• C. $15\text{m/s}$
• D. $10\text{m/s}$

Answer 1

The correct option is A $20\sqrt{3}\text{m/s}$

We know that velocity at the maximum height $=u\cos \theta$

According to question,

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

$\because \theta ={60}^{\circ }$

Range of a projectile $=R=\frac{u^2\sin 2\theta }{g}$

$R=\frac{u^2\sin \left({120}^{\circ }\right)}{g}=\frac{400\times \frac{\sqrt{3}}{2}}{10}=20\sqrt{3}\text{m/s}$

Hence, option $\left(A\right)$ is the correct answer.