Question

A basketball player is standing on the floor 10.0 m from
the basket as in Figure P4.50. The height of the basket is
3.05 m, and he shoots the ball at a 40.0o angle with the
horizontal from a height of 2.00 m. (a) What is the acceleration
of the basketball at the highest point in its trajectory?
(b) At what speed must the player throw the basketball so
that the ball goes through the hoop without striking the
backboard?

Answer 1

b. Dx = Vo^2*sin(2A)/g = 10 m
Vo^2*sin80/9.8 = 10
0.1Vo^2 = 10
Vo^2 = 100
Vo = 10 m/s[40o].