Question

The velocity at the maximum height of a projectile is half of its initial velocity $u$. Its range on the horizontal plane is:

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

Answer 1

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

Velocity at maximum height = half of initial velocity.
Velocity at maximum height = horizontal component.
$V_x=\frac{u}{2}$

$⟹u_y=\sqrt{u^2-\frac{u^2}{4}}=\frac{\sqrt{3}\times u}{2}$
i.e. $tan\theta =\sqrt{3}$

$\theta ={60}^{\circ }$
Range of projectile = $\frac{u^{2}sin2\theta }{g}=\frac{u^{2}sin{120}^{\circ }}{g}=\frac{\sqrt{3}{u}^2}{2g}$
Hence option B is correct.