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unit 5 review
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question 8
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the equation $h(t) = -16t^2 + 67t$ models the height $h$ in feet reached in $t$ seconds by a model rocket propelled straight up from the ground. when will the rocket be at a height of 63 feet?
seconds round to two decimal places.
basic funcs trig
enter a list of mathematical expressions more..

Explanation:

Step1: Set up the equation

We know \( h(t) = 63 \), so substitute into the function \( h(t)=- 16t^{2}+67t\):
\(63=-16t^{2}+67t\)
Rearrange it to the standard quadratic form \(ax^{2}+bx + c = 0\):
\(16t^{2}-67t + 63=0\)
Here, \(a = 16\), \(b=-67\), \(c = 63\)

Step2: Use the quadratic formula

The quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)
First, calculate the discriminant \(\Delta=b^{2}-4ac\)
\(\Delta=(-67)^{2}-4\times16\times63\)
\(= 4489-4032\)
\(=457\)

Then, find \(t\):
\(t=\frac{67\pm\sqrt{457}}{32}\)

Calculate the two solutions:
For the plus sign: \(t_{1}=\frac{67 + \sqrt{457}}{32}\approx\frac{67+21.38}{32}=\frac{88.38}{32}\approx2.76\)
For the minus sign: \(t_{2}=\frac{67-\sqrt{457}}{32}\approx\frac{67 - 21.38}{32}=\frac{45.62}{32}\approx1.43\)

Answer:

The rocket will be at a height of 63 feet at approximately \(t\approx1.43\) seconds and \(t\approx2.76\) seconds. If we consider the two - decimal - place rounded values, the answers are \(1.43\) and \(2.76\).