Chapter 13.3, Problem 34PPS

a.

To determine

To find: How far will the basketball travel.

Answer to Problem 34PPS

Distance of basketball shot $=R=9$ feet.

Explanation of Solution

Given information:

  $R=\frac{V_0{}^2\sin \left(2\theta \right)}{32}$

Where

  $R=$ Distance of a basketball shot

  $V_0=$ Initial velocity $=24$ feet per second

  $\theta =$ Angle with the ground $={75}^0$

Given

  $\theta ={75}^0$

  $V_0=24$ feet per second

On substituting values

We get

  $\begin{array}{l}R=\frac{V_0{}^2\sin \left(2\theta \right)}{32}\\ \Rightarrow R=\frac{{\left(24\right)}^2\sin \left(2\left(75\right)\right)}{32}\\ \Rightarrow R=\frac{576\left(0.5\right)}{32}\\ \Rightarrow R=\frac{288}{32}\\ \Rightarrow R=9\end{array}$

Therefore,

Distance of basketball shot $=R=9$ feet.

b.

To determine

To find: Initial velocity of the basketball.

Answer to Problem 34PPS

Initial velocity of basketball $=V_0=20.5$ feet per second.

Explanation of Solution

Given information:

  $R=\frac{V_0{}^2\sin \left(2\theta \right)}{32}$

Where

  $R=$ Distance of a basketball shot $=10$ feet

  $V_0=$ Initial velocity

  $\theta =$ Angle with the ground $={65}^0$

Given

  $\theta ={65}^0$

  $R=10$ feet

On substituting values

We get

  $\begin{array}{l}R=\frac{V_0{}^2\sin \left(2\theta \right)}{32}\\ \Rightarrow 10=\frac{V_0{}^2\sin \left(2\left(65\right)\right)}{32}\\ \Rightarrow 10=\frac{V_0{}^2\sin \left({130}^0\right)}{32}\\ \Rightarrow V_0{}^2\sin \left({130}^0\right)=32\times 10\\ \Rightarrow V_0{}^2\left(0.77\right)=320\\ \Rightarrow V_0{}^2=\frac{320}{0.77}\\ \Rightarrow V_0{}^2=415.6\\ \Rightarrow V_0=20.5\end{array}$

Therefore,

Initial velocity of basketball $=V_0=20.5$ feet per second.

c.

To determine

To find: At what angle was it shot.

Answer to Problem 34PPS

Angle $=\theta ={12.6}^0$ .

Explanation of Solution

Given information:

  $R=\frac{V_0{}^2\sin \left(2\theta \right)}{32}$

Where

  $R=$ Distance of a basketball shot $=10$ feet

  $V_0=$ Initial velocity

  $\theta =$ Angle with the ground $={65}^0$

Given

  $V_0=30$ feet per second

  $R=12$ feet

On substituting values

We get

  $\begin{array}{l}R=\frac{V_0{}^2\sin \left(2\theta \right)}{32}\\ \Rightarrow 12=\frac{{\left(30\right)}^2\sin \left(2\theta \right)}{32}\\ \Rightarrow 12=\frac{\left(900\right)\sin \left(2\theta \right)}{32}\\ \Rightarrow 900\sin \left(2\theta \right)=32\times 12\\ \Rightarrow \sin \left(2\theta \right)=\frac{384}{900}\\ \Rightarrow \sin \left(2\theta \right)=0.4266\\ \Rightarrow 2\theta ={\sin }^{-1}\left(0.4266\right)\\ \Rightarrow 2\theta ={25.3}^0\\ \Rightarrow \theta =\frac{{25.3}^0}{2}\\ \Rightarrow \theta ={12.6}^0\end{array}$

Therefore,

Angle $=\theta ={12.6}^0$ .