Chapter 6.3, Problem 43E

Solution

a)

To find: how many seconds after the ball is hit does it hit the wall.

The ball will hit the wall approximately in $2.8$ seconds.

Given information:

  $x=\left(150\cos 18°\right)t$ and $y=-16t^2+\left(150\sin 18°\right)t+3$ .

Formula used:

  $x=\left(v_0\cos \theta °\right)t$ and $y=-16t^2+\left(v_0\sin \theta °\right)t+h$ .

Calculation:

Substitute $400$ for $x$ in $x=\left(150\cos 18°\right)t$ .

  $\begin{array}{l}400=\left(150\cos 18°\right)t\\ 400=\left(150\right)\left(0.9511\right)t\\ 400=142.665t\\ \frac{400}{142.665}=t\\ t\approx 2.8\end{array}$

Hence, the ball will hit the wall approximately in $2.8$ seconds.

b)

To find: how high up the wall does the ball hit.

The ball will hit the wall at $7.34$ ft.

Given information:

  $x=\left(150\cos 18°\right)t$ and $y=-16t^2+\left(150\sin 18°\right)t+3$ .

Formula used:

  $x=\left(v_0\cos \theta °\right)t$ and $y=-16t^2+\left(v_0\sin \theta °\right)t+h$ .

Calculation:

Substitute $2.8$ for $t$ in $y=-16t^2+\left(150\sin 18°\right)t+3$ .

  $\begin{array}{l}y=-16{\left(2.8\right)}^2+\left(150\sin 18°\right)2.8+3\\ y=-16\left(7.84\right)+150\left(0.309\right)\left(2.8\right)+3\\ =-125.44+129.78+3\\ =7.34\end{array}$

Hence, the ball will hit the wall at $7.34$ ft.

c)

To explain: why Kevin’s hit might me caught by an outfielder and also explain why his would probably be not caught if it were hit at an angle of $20°$ with the horizontal.

The outfielder would not be able to catch a ball which would be at a height of $21.2$ ft when hit with an angle of $20°$ .

Given information:

  $x=\left(150\cos 18°\right)t$ and $y=-16t^2+\left(150\sin 18°\right)t+3$ .

Formula used:

  $x=\left(v_0\cos \theta °\right)t$ and $y=-16t^2+\left(v_0\sin \theta °\right)t+h$ .

Calculation:

Substitute $2.8$ for $t$ and $18°$ for $\theta$ in $y=-16t^2+\left(150\sin \theta \right)t+3$ .

  $\begin{array}{l}y=-16{\left(2.7\right)}^2+\left(150\sin 18°\right)2.7+3\\ =-16\left(7.29\right)+150\left(0.309\right)\left(2.7\right)+3\\ =-121+127.46+3\\ =7.35\end{array}$

Since the height of the ball is $7.35$ feet at $2.8$ seconds the outfielder might be able to catch it.

Substitute $2.8$ for $t$ and $20°$ for $\theta$ in $y=-16t^2+\left(150\sin \theta \right)t+3$ .

  $\begin{array}{l}y=-16{\left(2.8\right)}^2+\left(150\sin 20°\right)2.8+3\\ =-16\left(7.84\right)+150\left(0.342\right)\left(2.8\right)+3\\ =-125.44+143.64+3\\ =21.2\end{array}$

Hence, the outfielder would not be able to catch a ball which would be at a height of $21.2$ ft when hit with an angle of $20°$ .