Question

A ball is thrown from a tower 42 m. high above the ground with a velocity of 270 m/sec directed at 28⁰ from the horizontal. How long will the ball hit the ground?

Answer 1

$v_y=v\sin 28°=127~\frac ms,$

$t_1=\frac {v_y} g=13~s,$

$h_1=\frac{v_y^2}{2g}=823~m,$

$t_2=\sqrt{\frac{2(h+h_1)}g}=13.3~s,$

$t=t_1+t_2=26.3~s.$