Question

The velocity at the maximum height of a projectile is $\sqrt{3}/2$ times of its initial velocity of projection which is $u=20{\text{ms}}^{-1}$. Find the range of projectile on the horizontal plane.
(Take $g=10{\text{m/s}}^2$)

• A. $15\text{m/s}$
• B. $9\text{m/s}$
• C. $10\sqrt{3}\text{m/s}$
• D. $20\sqrt{3}\text{m/s}$

Answer 1

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

Given: $u=20\text{m/s}$

We know that velocity at the maximum height is

$u_{net}=u\cos \theta$

Because at maximum height, only horizontal component is present.

According to question,

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

$\Rightarrow \theta ={30}^{\circ }$

Range of a projectile is given by

$R=\frac{u^2\sin 2\theta }{g}=\frac{\left(20{)}^2\sin \right({60}^{\circ })}{10}$

$R=\frac{400\times \frac{\sqrt{3}}{2}}{10}=20\sqrt{3}\text{m}$

Hence, option $\left(D\right)$ is correct.