Question 502442: If an object is propelled upward from ground level with an initial velocity of 128 feet per second, it’s height s (in feet) is given by the formula (with t in seconds): s= -16t^2+128t
What is the maximum height the object will achieve?
At what time will this height be achieved?
At what time will the object return to the ground?
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! If an object is propelled upward from ground level with an initial velocity of 128 feet per second, it’s height s (in feet) is given by the formula (with t in seconds): s= -16t^2+128t
What is the maximum height the object will achieve?
At what time will this height be achieved?
At what time will the object return to the ground?
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The maximum height will be attained when the derivative ds/dt = 0
ds/dt = 0 = 128 - 32t
Solve for t:
32t = 128
t = 4 sec
So the maximum height = s(4) = 128*4 - 16*16 = 256 ft
By symmetry, we know that the time required to return to the ground will also be 4 sec, so the round trip will be 8 sec.
In mathematical terms, the two times when the object is at ground level will be the zeros of the height function
t(128 - 16t) = 0
This gives t = 0 (obviously) and t = 128/16 = 8
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