A ball is projected horizontally from the top of tower of height $80m$, with a velocity of $30m/s$. What is the magnitude of the velocity of the ball when it strikes the ground ? $(g=10m/s^2)$

The correct option is C $50m/s$
As the acceleration due to gravity is $vertical$ so the horizontal component $V_x=30m/s$ will remain $constant.$

and the vertical component will get increased from $0$ to $v$ given by $\sqrt{2gh}=\sqrt{2\times 10\times 80}=40m/s$

so the net velocity will be $V=\sqrt{v^2+v_x^2}=\sqrt{1600+900}=50m/s$