Question

A basketball player is 10 ft horizontally from the center of the basket, which is 10 ft off the ground. He is guarded by a good jumper. At what angle should the player aim the ball if it is thrown from a height of 7.1 ft with a speed of 26 ft/s? Use g = 32 ft/s². [Hint: 1/cos²? = 1 + tan²?]

? = ?

Answer 1

If we look at the problem from mathematical point of view you have to select angle A, so that horizontal velocity VcosA and vertical velocity VsinA became such that after time t, horizontal displacement Vt cosA=10ft and vertical VtsinA - 1/2 gt^2 = 10 - 7.6 = 2.4 ft. We get 2 equations:
26t cosA = 10
26t sinA - 16t^2 = 2.4 or 26t sinA = 2.4 + 16t^2
Square these equations and add them
676t^2(cos^2 A + sin^2 A) = 324 + 5.76 + 76.8 t^2 + 256 t^4
256 t^4 - 599.2 t^2 + 329.76 = 0
Substitute u=t^2
256 u^2 - 599.2 u + 329.76 = 0
Solve quadratic equation over u, select the largest solution and find cosA = 9/(13t) = 9/(13Sqrt(u)).