Answer:
49.2°
Explanation:
Using the equation of the trajectory
y = yā + (x - xā) tanĪø - ( g(x-xā)²) / 2( vācosĪø)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 Ć 4.57²) sec²θ / 2 Ć7.15²
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 ± ā ( b² - 4ac) / 2a
(- (- Ā 4.57) ± ā( (-4.57)² - (4 Ć 2 Ć 2.61))) / 2Ć 2
1.16 or 1.125
tanĪø = 1.16
Īø = tanā»Ā¹1.16 Ā = Ā 49.2°