Question

An object is launched at an upward velocity of 96 feet per second from the top of a building 160 feet above the ground. Use the vertical motion model h = - 16t² + vt + c where h is the ending height of the object, c is the initial height, in feet, of the object, t is the time in seconds, and v is the velocity of the object. How long will it take the object to reach the maximum height and what is the maximum height? Fill in the blanks with the appropriate values.​

Answer 1

$~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&96\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&160\\ \qquad \textit{of the object}\\ h=\textit{object's height}&h\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ h(t)=-16t^2+96t+160$

Check the picture below.

$\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-16}x^2\stackrel{\stackrel{b}{\downarrow }}{+96}x\stackrel{\stackrel{c}{\downarrow }}{+160} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 96}{2(-16)}~~~~ ,~~~~ 160-\cfrac{ (96)^2}{4(-16)}\right) \implies \left( - \cfrac{ 96 }{ -32 }~~,~~160 - \cfrac{ 9216 }{ -64 } \right)$

$(3~~,~~160 + 144)\implies \underset{~\hfill feet}{\stackrel{seconds~\hfill }{(\stackrel{\downarrow }{3}~~,~~\underset{\uparrow }{304})}}$