Question

1. marcus shoots an arrow straight upward. the velocity of the arrow is 160 ft/sec. the function that models the height of the arrow is ( f(x)=-5 + 160x-16x^{2} ), where ( x ) is the number of seconds that have passed since the arrow was shot. use your graphing calculator to answer the following questions. (a) how high is the arrow after 5 sec? (b) what is the maximum height that the arrow reaches before it falls back to the earth? (c) after how many seconds did the arrow reach the ground?

Explanation:

Step1: Solve part (a)

Substitute \(x = 5\) into the function \(f(x)=-5 + 160x-16x^{2}\).

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LATEXBLOCK0
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Step2: Solve part (b)

The function \(y = ax^{2}+bx + c\) (\(a=-16\), \(b = 160\), \(c=-5\)) has its vertex at \(x=-\frac{b}{2a}\).

$$x=-\frac{160}{2\times(-16)}=\frac{-160}{-32} = 5$$

Substitute \(x = 5\) into \(f(x)\):

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LATEXBLOCK1
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Step3: Solve part (c)

Set \(f(x)=0\), so \(-16x^{2}+160x - 5=0\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) (\(a=-16\), \(b = 160\), \(c=-5\))

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LATEXBLOCK2
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We get \(x_{1}=\frac{-160 + 159}{-32}=\frac{-1}{-32}\approx0.03\) and \(x_{2}=\frac{-160-159}{-32}=\frac{-319}{-32}\approx10\)

Answer:

a. \(395\) ft
b. \(395\) ft
c. Approximately \(10\) seconds