Question

The speed of a projectile at max height is half of its initial speed tge angle of projection is

Answer 1

u cos theta= u/2
cos theta= 1/2
theta= 60 as theta belongs to[0,90].

Answer 2

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

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

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

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