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?
Physics - drwls, Wednesday, February 6, 2013 at 9:56am
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
Physics - x, Friday, September 2, 2016 at 3:42pm