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°

step-by-step explanation:

we are given

we have to simplify it

firstly, we will factor it

now, we can cancel like terms

now, we can combine it

so, we get

good looking response