NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t) = -4.9t^{2} + 358t + 129 Assuming rocket launched at t = 0 and height is measured above sea level in meters.

asked 2021-01-05
NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by \(\displaystyle{h}{\left({t}\right)}=-{4.9}{t}^{{{2}}}+{358}{t}+{129}\) Assuming rocket launched at t = 0 and height is measured above sea level in meters.

Answers (1)

Sub part a) Rocket splashes into ocean Height will be zero \(\displaystyle-{4.9}{t}^{{{2}}}+{358}{t}+{129}={0}\) For a quadratic equation \(\displaystyle{a}{x}^{{{2}}}+{b}{x}+{c}={0}\), The solutions are, \(\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{\left({b}^{{{2}}}-{4}{a}{c}\right)}}}}}{{2}}{a}\) By substituting \(\displaystyle{a}=-{4.9},{b}={358},{c}={129}\)
\(\displaystyle{t}=-{0.36},{73.42}\) Since time cannot be negative. Hence the rocket splashes down into ocean after 73.42 seconds (rounded to two decimals). Sub part b) The rocket reaches to peak when the first derivative of the height equation is zero. \(\displaystyle\Rightarrow{h}'{\left({t}\right)}={0}\)
\(\displaystyle\Rightarrow-{9.8}{t}+{358}={0}\) Formula : derivative of \(\displaystyle{x}^{{{n}}}={n}{x}^{{{n}-{1}}}\)
\(\displaystyle\Rightarrow{t}=\frac{{358}}{{9.8}}={36.53}\) (Rounded to two decimals) \(\displaystyle{h}'{\left({36.53}\right)}=-{9.8}{\left({36.53}\right)}+{358}\)
\(\displaystyle\Rightarrow{h}{\left({t}\right)}\) has its peak value at t=36.53 seconds and it's value is, \(\displaystyle{h}{\left({36.53}\right)}=-{4.9}{\left({36.53}\right)}^{{{2}}}+{358}{\left({36.53}\right)}+{129}\)
PSK= 6,667.98 Hence the peak value of rocket is at 36.53 seconds and the height is 6667.98 meters.

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