nasa launches a rocket at t = 0 seconds. suppose its height, in meters above sea - level, as a function of…

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.\nhow high above sea - level does the rocket get at its peak? round your answer to 2 decimal places.\nthe rocket peaks at 2847.18 × meters above sea - level.

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.\nhow high above sea - level does the rocket get at its peak? round your answer to 2 decimal places.\nthe rocket peaks at 2847.18 × meters above sea - level.

Answer

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