A basketball player throws the ball at a 36° angle above the horizontal to a hoop which is located a horizontal distance L = 8.2 m from the point of release and at a height h = 0.2 m above it. What is the required speed if the basketball is to reach the hoop?

Answer 1

Let t be the time of flight

Vcos32*t = 8.2
Vsin32*t -4.9 t^2 = 0.2

0.848 V*t = 8.2
0.530 V*t -4.9 t^2 = 0.2

First eliminate V by substitution and solve the resulting equation for t.

5.125 -4.9t^2 = 0.2
4.9t^2 = 4.925
t = 1.003 s
V = 8.2/(.848*1.003) = 9.64 m/s

Answer 2

x+y