A basketball player is trying to make a half-court jump shot and releases the ball at the height of the basket. Assuming that the ball is launched at 51.0°, 14.0m from the basket, what speed must the player give the ball?

Nov 22, 2015

$11.83 \text{m/s}$

Explanation:

The range of the basketball is given by:

$d = \frac{{v}^{2}}{g} \sin 2 \theta$

$\therefore {v}^{2} = \frac{d \text{g}}{\sin 2 \theta}$

${v}^{2} = \frac{14 \times 9.8}{0.981}$

$v = 11.83 \text{m/s}$