Question 165194: 22.) Find he exact solution.
The height in feet for a ball thrown upward at 48 feet per second is given by
s(t)= - 16r^2 + 48t. Where t is the time in seconds after the ball is tossed. What is the maximum height that the ball will reach?
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! s(t)= - 16t^2 + 48t
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The idea is that the equation describes a parabola. If the coefficient 'a' is negative, the parabola opens downward. Thus the vertex will give you the "maximum".
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The vertex form of a parabola is:
y= a(x-h)^2+k
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The idea is to convert the given equation into the above form.
s(t)= - 16t^2 + 48t
s(t)= -16(t^2 - 4t)
Completing the square, we have:
s(t)= -16(t^2 - 4t + 4) + 64
s(t)= -16(t-2)^2 + 64
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We now "see" that the vertex is at:
(h,k) = (2, 64)
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Which says, that at
2 seconds, the ball will reach the maximum height of 64 feet.
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