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
v3 u2
29
(2) V3 u?
g
2u?
(4)​

Answer 1

At maximum height, only horizontal velocity exists. Thus,

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

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

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

Then, horizontal range,

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

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

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