If a ball is thrown into the air with a velocity of 80 ft/s, its height in feet after
t seconds is given by s(t) = 80t - 16t
2
. It will be at maximum height when its
instantaneous velocity is zero. Find its average velocity from the time it is
thrown (t = 0) to the time it reaches its maximum height.

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\[80t-16t ^{2}\]

so i have \[\frac{(80(2+h)-16(2+h)^{2}-(80(2)-16(2)^{2})}{ h }\]

when i work the problem this way i don't get the right answer.

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