The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.

49.2°

Explanation:

Using the equation of the trajectory

y = y₀ + (x - x₀) tanθ - (g (x-x₀) ²) / 2 (v₀cosθ) ²

1 / cos θ = sec θ

y - y₀ = (x - x₀) tanθ - (g (x-x₀) ²) sec²θ / 2v²

y - y₀ = 3.05 m - 2.44 m (height of the basket - height the thrower released the ball) = 0.61 m

x - x₀ = 4.57

g = 9.81 m/s²

v = 7.15 m/s

0.61 m = 4.57 tanθ - (9.81 * 4.57²) sec²θ / 2 * 7.15²

0.61 m = 4.57 tanθ - 2 sec²θ

but sec²θ = tan²θ + 1

0.61 m = 4.57 tanθ - 2 (tan²θ + 1)

0.61 m = 4.57 tanθ - 2tan²θ - 2

0 = - 2tan²θ + 4.57 tanθ - 2 - 0.61 m

0 = - 2tan²θ + 4.57 tanθ - 2.61

multiply by - 1 both side

0 = 2 tan²θ - 4.57 tanθ + 2.61

let x = tan θ

2 x² - 4.57 x + 2.61

using quadratic formula

-b ± √ (b² - 4ac) / 2a

( - ( - 4.57) ± √ ((-4.57) ² - (4 * 2 * 2.61))) / 2 * 2

1.16 or 1.125

tanθ = 1.16

θ = tan⁻¹1.16 = 49.2°