Answer to Question #185841 in Classical Mechanics for ddd

Question

The boy in question 1 now throws his ball horizontally out of a high window with a velocity of 3.1 m s-1. How long will it take to reach the ground 18m below?

Answer 1

$v_x=3 .1~\frac ms,~v_y=0~\frac ms,$

$h=v_yt+\frac{gt^2}{2}=\frac{gt^2}{2}\implies$

$t=\sqrt{\frac{2h}{g}}=1.9~s.$