A basketball player throws the ball at a 53degree angle above the horizontal to a hoop which is located a horizontal distance L = 2.4 m from the point of release and at a height h = 0.9 m above it.What is the required speed if the basketball is to reach the hoop?

Answer 1

h = 0.5g*t^2 = 0.9 m.
4.9*t^2 = 0.9
t^2 = 0.1837
Tf = 0.429 s. = Fall time.

Tr = Tf = 0.429s. = Rise time.

Y = Yo + g*Tr = 0
Yo = -g*Tr = 9.8 * 0.429 = 4.2 m/s.

Yo = Vo*Sin53 = 4.2
Vo = 4.2/Sin53 = 5.26 m/s. = The required speed.