Answer:d-The thrown ball
Explanation:
Given
First ball is dropped from window at a height suppose h
therefore velocity of ball at the ground
[tex]v=\sqrt{2gh}[/tex]
second ball is thrown horizontally suppose with a velocity of u therefore taking horizontal and vertical motion separately
horizontal velocity will remain same as there is no gravity in horizontal direction
vertical velocity at ground of second ball
[tex]v_y=\sqrt{2gh}[/tex]
horizontal velocity is u
net velocity will be
[tex]v_{net}=\sqrt{v_y^2+u^2}[/tex]
[tex]v_{net}=\sqrt{2gh+u^2}[/tex]
thus velocity of second ball be more as compared to first