Question 316746
1.{{{h(2)= -16(2)^2 +80(2)+ 50}}}
{{{h(2)=-64+160+50}}}
{{{highlight( h(2)=146)}}}
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2.{{{h(5)= -16(5)^2 +80(5)+ 50}}}
{{{h(5)=-400+400+50}}}
{{{highlight(h(5)=50)}}}
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3.Complete the square to get to vertex form.
The max. height occurs at the vertex.
 {{{h(t)=-16t^2+80t+50}}}
{{{h(t)=-16(t^2-5t+25/4)+50+100}}}
{{{h(t)=-16(t-5/2)^2+150}}}
(5/2,150) is the vertex.
{{{highlight(hmax=150)}}}
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4. {{{h(t)=-16(t-5/2)^2+150}}}
{{{-16(t-5/2)^2+150=0}}}
{{{16(t-5/2)^2=150}}}
{{{(t-5/2)^2=75/8}}}
{{{t-5/2=0 +- sqrt(75/8)}}}
{{{t=5/2 +- (5/2)*sqrt(3/2)}}}
The time we need is the positive solution,
{{{t=5/2+(5/2)*sqrt(3/2)}}}
{{{highlight(t=5/2+(5/4)*sqrt(6))}}} or approximately
{{{t=5.56}}} sec.
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{{{ graph( 300, 300, -2, 8, -10, 190, -16x^2 +80x + 50) }}}