A body of mass $1kg$ is projected from ground at an angle ${30}^{\circ }$ with horizontal on a level ground at a speed $50m/s$. The magnitude of change in momentum of the body during its flight is $(g=10m/s^2)$

The correct option is A. $50kgm{s}^{-1}$.

We know that in the horizontal plane, the initial velocity of projectile = final velocity of the projectile

$\overrightarrow{p_i}=mu\cos \theta \hat{i}+mu\sin \theta \hat{j}$

$\overrightarrow{p_f}=mu\cos \theta \hat{i}-mu\sin \theta \hat{j}$

$\Delta \overrightarrow{p}=\overrightarrow{p_f}-\overrightarrow{p_i}=-2mu\sin \theta \hat{j}$

$|\Delta \overrightarrow{p}|=2mu\sin \theta$

$=2\times 1\times 50\times \frac{1}{2}=50kgm{s}^{-1}$

Alternative: In the horizontal direction momentum does not change,
$\Delta P_x=0$
In vertical direction
$\Delta P_y=\left(mu\sin \theta \right)-(-mu\sin \theta )$ (Take downward as positive)
$=2mu\sin \theta$ downward
$=2\times 1\times 50\times \frac{1}{2}=50kgm{s}^{-1}$