Question

A baseball is hit, following a path represented by x = 130t and y = 3.2 + 42t − 16t 2 for 0 ≤ t ≤ 3.

Answer 1

Part A)

Evaluate x(t) and y(t) at t=0.2, 1.2 and 2.4 to find the ordered pairs.

$\begin{gathered} x=130t \\ y=3.2+42t-16t^2 \end{gathered}$

t=0.2

$\begin{gathered} x=130\times0.2=26 \\ y=3.2+42\times0.2-16\times0.2^2=10.96 \end{gathered}$

Then, the ordered pair for t=0.2 is (26,10.96).

t=1.2

$\begin{gathered} x=130\times1.2=156 \\ y=3.2+42\times1.2-16\times1.2^2=30.56 \end{gathered}$

Then, the ordered pair for t=0.2 is (156,30.56).

t=2.4

$\begin{gathered} x=130\times2.4=312 \\ y=3.2+42\times2.4-16\times2.4^2=11.84 \end{gathered}$

Then, the ordered pair for t=2.4 is (312,11.84).

Part B)

Find a rectangular equation (y as a function of x) to find the height of the ball when it reaches a horizontal distance of 320ft. To do so, isolate t from the equation for x:

$\begin{gathered} x=130t \\ \Rightarrow t=\frac{x}{130} \end{gathered}$

Replace t=x/130 into the equation for y:

$\begin{gathered} y=3.2+42t-16t^2 \\ \Rightarrow y=3.2+42(\frac{x}{130})-16(\frac{x}{130})^2 \\ \Rightarrow y=3.2+\frac{42}{130}x-16\times\frac{x^2}{16,900} \\ \Rightarrow y=3.2+\frac{42}{130}x-\frac{16}{16,900}x^2 \\ \Rightarrow y=3.2+\frac{42}{130}x-\frac{4}{4,225}x^2 \end{gathered}$

Replace x=320 to find the height of the ball:

$y=3.2+\frac{42}{130}(320)-\frac{4}{4225}(320)^2=9.6378...$

Since the height of the ball is less than the height of the fence when it reaches a horizontal distance of 320ft, then the baseball doesn't travel over the fence.

Part C)

A rectangular equation to represent the plane curved was already found in Part B:

$y=3.2+\frac{42}{130}x-\frac{4}{4225}x^2$