Question

A rocket is fired upward from some initial distance above the ground. Its height in​ feet, h,above the​ ground, t seconds after it is​ fired, is given by h=−16t^2+96t+2560.

What is the rocket's maximum height?

How long does it take for the rocket to reach its max height?

After it's fired, the rocket reaches the ground at t=__ sec.

Answer 1

check the picture below.

`\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-16}x^2\stackrel{\stackrel{b}{\downarrow }}{+96}x\stackrel{\stackrel{c}{\downarrow }}{+2560} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 96}{2(-16)}~~~~ ,~~~~ 2560-\cfrac{ (96)^2}{4(-16)}\right)\implies \left(-\cfrac{96}{-32} ~~,~~2560-\cfrac{9216}{-64} \right)`

`\left( 3~~,~~2560+144 \right)\implies \underset{maximum~height}{\stackrel{\textit{it took this long}}{(\stackrel{\downarrow }{3}~~,~~\underset{\uparrow }{2704})}} \\\\[-0.35em] \rule{34em}{0.25pt}`

`\bf h(t)=-16t^2+96t+2560\implies 0=-16t^2+96t+2560 \\\\\\ 0=-16(t^2-6t-160)\implies 0=t^2-6t-160 \\\\\\ 0=(t-16)(t+10)\implies t= \begin{cases} \boxed{16}\\ -10 \end{cases}`

"t" is an amount in seconds, for this specific phenomena, it cannot be negative, so -10 is not a feasible value.