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A given projectile has a height function given by h(t)=-16(t-4)^2+400. what is its maximum height and at what time, t, does it occur?

A given projectile has a height function given by h(t)=-16(t-4)^2+400. what is its-example-1
User Yoram
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1 Answer

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Check the picture below.

well, since h(t) is already in vertex form, well hell we pretty much can see the vertex right off


~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{

now, when does it start above the ground?

well, from the picture below, notice, the initial height happens at 0 seconds, namely when t = 0, well, let's set it to 0.


h(t)=-16(t-4)^2 + 400\implies h(0)=-16(0-4)^2 + 400 \\\\\\ h(0)=-16(4)^2 + 400\implies h(0)=-256+400\implies \stackrel{initial~height}{h(0)=144}

A given projectile has a height function given by h(t)=-16(t-4)^2+400. what is its-example-1
User Liwen
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