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.

Respuesta :

Answer:

49.2°

Explanation:

Using the equation of the trajectory

y = yā‚€ + (x - xā‚€) tanĪø - ( g(x-xā‚€)²) / 2( vā‚€cosĪø)²

1 / cos Īø = sec Īø

y - yā‚€ = (x - xā‚€) tanĪø - ( g(x-xā‚€)²) sec²θ / 2v²

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

x - xā‚€ = 4.57

g = 9.81 m/s²

v = 7.15 m/s

0.61 m = Ā 4.57 tanĪø - (9.81 Ɨ 4.57²) sec²θ / 2 Ɨ7.15²

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 Īø

2 x²  -  4.57 x + 2.61

using quadratic formula

-b ± √ ( b² - 4ac) / 2a

(- (- Ā 4.57) ± √( (-4.57)² - (4 Ɨ 2 Ɨ 2.61))) / 2Ɨ 2

1.16 or 1.125

tanĪø = 1.16

θ = tan⁻¹1.16  =  49.2°