Chapter 6.3, Problem 45E

Solution

a.

To check: Whether the ball clear the fence if there is $5$ -ft/sec split-second wind gust.

Yes, the ball clears the fence if there is $5$ -ft/sec split-second wind gust.

Given information: It is given that split-second wind gust is $5$ -ft/sec. The direction of the wind is horizontal out with the ball. The height of the fence is $30$ -ft.

Calculation:

The parametric equation obtained with velocity $120$ ft/sec, angle $30°$ and the height of ball is above $4$

ft from the ground is given by:

  $\begin{array}{l}x=120\cos 30°\\ y=-16t^2+\left(12\sin 30°\right)t+4\end{array}$

The horizontal speed of the ball will be affected by the wind gust. Add $5$ to the speed of the ball.

  $x=\left(120\cos 30°+5\right)t$   ...... (1)

Consider the same equation for $y$ .

  $y=-16t^2+\left(12\sin 30°\right)t+4$   ...... (2)

The height of the ball from home plate is $350$ feet. Substitute $350$ for $x$ in the equation (1).

  $\begin{array}{c}350=\left(120\cos 30°+5\right)t\\ t=\frac{350}{120\cos 30°+5}\end{array}$

Substitute $\frac{350}{120\cos 30°+5}$ for $t$ in the equation (2).

  $\begin{array}{c}y=-16{\left(\frac{350}{120\cos 30°+5}\right)}^2+\left(12\sin 30°\right)\left(\frac{350}{120\cos 30°+5}\right)+4\\ \approx 31.59\end{array}$

It is given that the height of the fence is $30$ -ft. Here, $30<31.59$ . So, ball will clear the fence.

Thus, the ball clears the fence.

b.

To find: The height by which ball clear the fence.

The ball clears the fence by the distance of $1.59$ ft.

Given information: It is given that split-second wind gust is $5$ -ft/sec. The direction of the wind is horizontal out with the ball. The height of the fence is $30$ -ft.

Calculation:

From part (a), the height of the ball when it is above the fence is $31.59$ ft. The height of the fence is $30$ -ft.

Subtract the height of the fence from the height of the ball.

  $31.59-30=1.59\text{ft}$

Thus, the ball clears the fence by the distance of $1.59$ ft.