Question

a women is standing behind a 4m high wall. she throws a ball with a velocity of 10.0m/s 60 degrees above the horizontal releasing it from a height of 2 meters. you are on the other side of the wall from the women at a considerable distance away observing what is happening for how long is the ball visible above the wall?

Answer 1

$$h(t)=h_0-0.5gt^2+v\sin60 t=H$$

$$2-0.5(9.8)t^2+10\sin60 t=4$$

$$t_1=0.27\ s,t_2=1.49\ s$$

Needed time:

$$t_2-t_1=1.49-0.27=1.22\ s$$