SOLUTION: Assume the acceleration of the object is a(t) = −32 feet per second per second. (Neglect air resistance.) A ball is thrown vertically upward from a height of 5 feet with an

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Question 1127310: Assume the acceleration of the object is a(t) = −32 feet per second per second. (Neglect air resistance.)
A ball is thrown vertically upward from a height of 5 feet with an initial velocity of 72 feet per second. How high will the ball go?
Found 2 solutions by stanbon, addingup:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Assume the acceleration of the object is a(t) = −32 feet per second per second. (Neglect air resistance.)
A ball is thrown vertically upward from a height of 5 feet with an initial velocity of 72 feet per second. How high will the ball go?
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Height equation::
h(t) = -32t^2+72t+5
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Maximum height occurs when t = -b/(2a) = -72/(2*-32) = 1.125
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h(1.125) = -32(1.125)^2 + 72(1.125)+5 = 45.5 ft
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Cheers,
Stan H.
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Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
using a(t) = -32ft/s^2
a(t) = -32ft/s^2 --> v(t) = ∫-32dt --> v(t) = -32t + C
v(0) = 72 --> v(0) = -32(0)+C = 72 so our velocity function is:
v(t) = -32t+72 --> at the maximum height the velocity will be zero, so we can restate:
0 = -32t+72 --> t = 2.25
Now, to find the maximum height
s(t) = ∫v(t)dt = ∫(-32t+ 72)dt =
sorry I've run out of time. Maybe the above will help you find the answer
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