Question

The speed of a projectile at its maximum height is half of its initial speed. the angle of projection is

Answer 1

at maximum height the vertical component of velocity is zero and horizontal component remains ucos theta.
so ucos theta=u/2
cos theta=1/2
theta=π/3

Answer 2

a particle is projected with speed u m/s which is inclined at an angle $\theta$ with horizontal.
initial velocity of particle , $\vec{u}=ucos\theta\hat{i}+usin\theta\hat{j}$

we know, at maximum height , only horizontal component of velocity exists.
so, velocity of particle at maximum height ,
$\vec{v}=ucos\theta\hat{i}$

a/c to question,
speed of project at its maximum height is half of its initial speed.
so, $2|\vec{v}|=|\vec{u}|$

$2ucos\theta=u$

$cos\theta=\frac{1}{2}$

$\theta=\frac{\pi}{3}$

hence, angle of projection is π/3