A basketball player throws the ball at a 45 angle above the horizontal to a hoop which is located a horizontal distance L = 2.6 from the point of release and at a height h = 0.6 above it. What is the required speed if the basketball is to reach the hoop?
time to cover the 2.6m horizontal distance: 2.6/(v cosθ) = 2.6√2/v
Now for the vertical distance:
v sinθ t - 4.9t^2 = 0.6
v/√2 (2.6√2/v) - 4.9(2.6√2/v)^2 = 0.6
v = 5.75 m/s