detailed solution

Correct option is q

There is no change in horizontal velocity, hence no change in momentum in horizontal direction. The vertical velocity at t=10sec is
So change in momentum in vertical direction is  $\mathrm{v}=38\times \sin {60}^0-\left(9.8\right)\times 10=-13.13\mathrm{m}/\sec$
So change in momentum in vertical direction is$=\left(0.5\times 98\times \sqrt{3}/2\right)-\left[-\left(0.5\times 13.13\right)\right]$
$=42.434+6.56=48.997\approx 49$

detailed solution

Correct answer is 49

There is no change in horizontal velocity, hence no change in momentum in horizontal direction. The vertical velocity at t=10sec is
So change in momentum in vertical direction is  $\mathrm{v}=38\times \sin {60}^0-\left(9.8\right)\times 10=-13.13\mathrm{m}/\sec$
So change in momentum in vertical direction is$=\left(0.5\times 98\times \sqrt{3}/2\right)-\left[-\left(0.5\times 13.13\right)\right]$
$=42.434+6.56=48.997\approx 49$