Question

During a basketball game, a player shoots a ball from half-court. When the ball reaches its maximum height of 4.5 m above the floor, it is moving at 5.0 m/s. If the ball was released from 2.5 m above the floor, what was the angle above the horizontal of the ball’s initial velocity?

Answer 1

51.38

Explanation

Step 1

Draw

at the point of maximum heigth, we have

$\begin{gathered} v_y=0 \\ v_x=v_0=5\text{ m/s} \\ H_{\max }=2\text{ m} \end{gathered}$

so, we can use the formula

$\begin{gathered} v_x=v_0\cdot\cos \emptyset \\ 5=v_0\cdot\cos \emptyset \\ \frac{5}{v_0}=\cos \emptyset \end{gathered}$

and

$\begin{gathered} y\text{ max} \\ h=\frac{v^2_0\sin ^2\emptyset}{2g} \\ \text{replace} \\ \frac{2\cdot2(9.8)}{\sin ^2\emptyset}=v^2_0 \\ v=\sqrt[]{\frac{2\cdot2(9.8)}{\sin^2\emptyset}}=\frac{5}{\cos o\text{ }\emptyset} \\ \\ \frac{6.2}{\sin o} \\ \frac{6.2}{5}=\frac{\sin \emptyset}{\cos \emptyset} \\ \text{1}.25=\tan g \\ angle=51.38 \end{gathered}$

I hope this helpsyou