A basketball is shot from 2 meters up at an angle of 60° above the x axis at an initial velocity of 9 m/s.What is the maximum height the ball will reach?

Answer 1

initial distance up = 2
initial velocity component up = 9 sin 60 = 7.79
v = 9 sin 60 - 9.8 t
when v = 0, we are there
9.8 t = 7.79
t = .795 seconds to top
h = 2 + 7.79(.795) - 4.9(.795^2)