Question

nasa launches a rocket at t = 0 seconds. suppose its height, in meters above sea - level, as a function of time is given by h = - 4.9t² + 163t + 269.
how high above sea - level does the rocket get at its peak? round your answer to 2 decimal places.
the rocket peaks at 2847.18 × meters above sea - level.

Explanation:

Step1: Find the time when the rocket peaks

For a quadratic function \(h(t)=at^{2}+bt + c\) (here \(a=-4.9\), \(b = 163\), \(c = 269\)), the time \(t\) at which the vertex (peak) occurs is given by the formula \(t=-\frac{b}{2a}\).
Substitute \(a=-4.9\) and \(b = 163\) into the formula:
\(t=-\frac{163}{2\times(-4.9)}=\frac{163}{9.8}\approx16.63\)

Step2: Find the height at the peak time

Substitute \(t = 16.63\) into the height function \(h(t)=-4.9t^{2}+163t + 269\)
\(h(16.63)=-4.9\times(16.63)^{2}+163\times16.63+269\)
First, calculate \((16.63)^{2}=276.5569\)
Then, \(-4.9\times276.5569=-1355.12881\)
\(163\times16.63 = 2710.69\)
\(h(16.63)=-1355.12881+2710.69+269\)
\(h(16.63)=1624.56119\approx1624.56\)

Answer:

\(1624.56\) meters