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An object is projected vertically upward from the top of a building with an initial velocity of 128 ft/sec.?
An object is projected vertically upward from the top of a building with an initial velocity of 128 ft/sec. Its distance s(t) in feet above the ground after t seconds is given by the equation
s(t) = −16t^2 + 128t + 110.
(a) Find its maximum distance above the ground.
Please help! I can't find this anywhere
2 Answers
- PolygonLv 79 years agoFavorite Answer
At max or min ds/dt = 0
-32t + 128 = 0
t = 4seconds
Substituting s(4) = -256 + 512 + 110 = 366feet
- Anonymous9 years ago
s = -16t² + 128t + 110
s = -16(t² - 8t + 16) + 110 + 256
s = -16(t - 4)² + 366
So after 4 seconds the object reaches a maximum height of 366 feet.