A basketball is thrown horizontally with an initial speed of 4.20 m/s. A straight line drawn from the point to the landing point makes an angle of 30.0 with the horizontal. what was the release height?

Answer 1

It falls a distance h with zero initial velocity down
so
h = (1/2) (9.8) t^2 = 4.9 t^2
tan 30 = h/d where d is horizontal distance to point hit on floor so d = h/.577 = 1.73 h
d = 4.2 t
so
4.2 t = 1.73 h
t = .412 h
then
h = 4.9 t^2 = 4.9 (.412^2)(h^2)
h = .831 h^2
h = 1.20 meters