Question

The free throw line in basketball is 4.570 m (15 ft) from the basket, which is 3.050 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.157 m/s, releasing it at a height of 2.440 m above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket?

Answer 1

Answer:

$\theta = 86.491^{\textdegree}$

Explanation:

The equations for the horizontal and vertical position of the ball are, respectivelly:

$4.570\,m = [(7.157\,\frac{m}{s})\cdot\cos \theta]\cdot t\\3.050\,m = 2.440\,m +[(7.157\,\frac{m}{s})\cdot \sin \theta]\cdot t - \frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}$

By isolating each trigonometric component and summing each equation:

$20.885\,m^{2} = [51.223\,\frac{m^{2}}{s^{2}}\cdot \cos^{2} \theta]\cdot t^{2}$

$[0.61\,m + \frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}]^{2} = [51.223\,\frac{m^{2}}{s^{2}}\cdot \sin^{2} \theta]\cdot t^{2}$

$21.257\,m^{2} + (5.982\,\frac{m^{2}}{s^{2}})\cdot t^{2}+(24.044\,\frac{m^{2}}{s^{4}} )\cdot t^{4} = (2623.796\,\frac{m^{2}}{s^{2}})\cdot t^{2}$

$21.257\,m^{2} - (2617.814\,\frac{m^{2}}{s^{2}})\cdot t^{2}+(24.044\,\frac{m^{2}}{s^{4}} )\cdot t^{4} = 0$

The positive real roots are:

$t_{1} = 10.434\,s,t_{2} = 0.09\,s$

The needed angle is:

$\theta = \cos^{-1} [\frac{4.570\,m}{(7.157\,\frac{m}{s} )\cdot t} ]\\\theta_{1} = 86.491^{\textdegree}\\\theta_{2} = NaN$