Question

A model rocket is launch from the ground with initial velocity of 352 ft/sec how long will it take the rocket to reach its maximum height show all work in the space provided

Answer 1

check the picture below. So the rocket reaches its height at its vertex.

the rocket is being launched from the ground, and therefor its initial height is 0, thus

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{352}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+352t+0$

so we can get the vertex coordinates by using its coefficients, how many "t" seconds will it take?  well, that'd be the x-coordinate of its vertex.

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+352}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{352}{2(-16)}~~,~~\qquad \right)\implies \left(\cfrac{352}{32}~~,~~\qquad \right)\implies (\stackrel{seconds}{11}~~,\qquad )$