Question

momentum at the highest point of trajectory of projectile

Answer 1

Let the initial velocity of the projectile be $u\ ms^{-1}$.

At the highest point,

$v_x=u\cos\theta\quad[a=0\ ms^{-2}]\\\\v_y=0\quad[u\sin\theta=gt]\\\\\therefore\ v=\sqrt{u^2\cos^2\theta+0^2}=u\cos\theta$

Hence the momentum is,

$\dfrac{1}{2}mv^2=\dfrac{1}{2}mu^2\cos^2\theta$