Answer to Question #250477 in Physics for Will

Question

A basketball is shot from an initial height of 2.40m with an initial speed of 15m/s directed at an angle of 35 degrees above the horizontal. How far from the basket (in meters) was the player if he made a basket

Answer 1

The range is

$$R=vT\cos\theta.$$

The time of flight upward:

$$t_1=\frac {v\sin\theta}g.$$

The time downward:

$$t_2=\sqrt{\frac{2h}g}=\sqrt{\frac{2(v^2\sin^2\theta/(2g)+h_0)}g}.$$

Range:

$$R=v\cos\theta(t_1+t_2)=\\\space\\ =v\cos\theta\bigg(\frac {v\sin\theta}g+\sqrt{\frac{2(v^2\sin^2\theta/(2g)+h_0)}g}\bigg)=\\\space\\ =24.58\text{ m}.$$