Asked by Anonymous

an arrow is shot into the air with an initial velocity of 96 feet per second. the height in feet of the arrow t seconds after it was shot into the air is given by the function h(x)=-16t^2+96t. find the maximum height of the arrow. what is the answer
Answered by Henry
h(x) = -16t^2 + 96t.

The parabola opens downward. Therefore,
the max. point is the vertex.

t(V) = -b/2a = -96 / -32 = 3s = The value of t at the vertex.

h=-16*3^2 + 96*3 = -144 + 288 = 144Ft







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Answered by Anonymous
how long was the arrow in the air
Answered by Anonymous
how long was the arrow in the air
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