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

Step-by-step 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 \\ (5)/(v_0)=\cos \emptyset \end{gathered}`

and

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

I hope this helpsyou