Abasketball player is 15 ft horizontally from the center of the basket, which is 10 ft off the ground. at what angle should the player aim the ball if it is thrown from a height of 8.2 ft with a speed of 29 ft/s ?

The direction of path should be inclined at and angle of
sin (theta) = (10 - 8.2)/ 15 = 3 / 25
this angle is the ratio of the vertical displacement of the projectile and the horizontal displacement of the projectile.time taken to reach the top most point can found out by by using the formula v = u + at in the vertical direction 
intial velocity in vertical direction = 29 sin (theta)
final velocity in vertical direction = 0 
acceleration in vertcal direction = - 32 ft / s^2   so u will get the time to be 29 sin (theta) / 32
vertical displacement = [29 sin (theta)] [(29 sin ( theta)/32)] - (1/2)32 [(29 sin (theta)/32)]^ 2 
horizontal displacement = [29 cos ( theta )] * 29 sin (theta) / 32
vertical displacement /   horizontal displacement=   3/25 
if u substitute the values u will get tan (theta) = 6 / 25

Answer: 900°

good luck on your studies! : )

david edward.