Question

PLEASE HELP ME! I really need help y'all...

An object is launched from a platform.
Its height (in meters), x seconds after the launch, is modeled by:

h(x)= -5(x-4)^2+180

What is the height of the object at the time of launch?

________ meters

Answer 1

It is 10

Explanation:

:)

Answer 2

just a quick clarification, tis usually -4.9 and that's a rounded number to reflect earth's gravity on an object in motion, but -5 is close enough :)

`\bf ~~~~~~\textit{initial velocity in meters} \\\\ h(t) = -4.9t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}\\ \qquad \textit{of the object}\\ h=\textit{object's height}\\ \qquad \textit{at`

`\bf h(x)=-5(\stackrel{\mathbb{F~O~I~L}}{x^2-8x+16})+180\implies h(x)=-5x^2+40x-80+180 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x)=-5x^2+\stackrel{\stackrel{v_o}{\downarrow }}{40} x+\stackrel{\stackrel{h_o}{\downarrow }}{\boxed{100}}~\hfill`