Question

The speed at the maximum height of a projectile is half of its initial speed of projection (u). the horizontal range of the projectile is

Answer 1

range is
$\frac{ \sqrt{3} \times {u}^{2} }{2g}$

Answer 2

At the topmost position, only horizontal velocity exists. Thus,

$\displaystyle\longrightarrow\sf{u\cos\theta=\dfrac{u}{2}}$

$\displaystyle\longrightarrow\sf{\cos\theta=\dfrac{1}{2}}$

$\displaystyle\longrightarrow\sf{\theta=60^o}$

Then, the horizontal range,

$\displaystyle\longrightarrow\sf{R=\dfrac{u^2\sin(2\theta)}{g}}$

$\displaystyle\longrightarrow\sf{R=\dfrac{u^2\sin120^o}{g}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{R=\dfrac{u^2\sqrt3}{2g}}}}$