SOLUTION: An obvject is thrown upward into the air with an initial velocity of 128 feet per second. The formula h(t)=128t-16t squared gives its hight above the ground after t seconds. What i

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Question 41288: An obvject is thrown upward into the air with an initial velocity of 128 feet per second. The formula h(t)=128t-16t squared gives its hight above the ground after t seconds. What is the height after 2 seconds. What is the maximum hight reached. For how many seconds will the object be in the air. After how many seconds will it reach the maximum hight. After how many seconds will it hit the ground. I could use some help. I would greatly apreciate it.
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
h%28t%29=128t-16t%5E2
When t=2, h%282%29=128%2A2-16%2A2%5E2 = 192
So the height reached after 2 seconds is 192 ft.

To find maximum height we shall first express h(t) as a difference of two quantities the last one being a function of 't'.
Let's see this.
h%28t%29=128t-16t%5E2
or h%28t%29=+16%2A4%5E2+-+16%28t%5E2+-+2%2At%2A4+%2B+4%5E2%29
or h%28t%29=+16%5E2+-+16%28t+-+4%29%5E2
Now, h(t) is maximum when the second quantity i.e. that one on the right side of minus sign must be minimum.
Now, %28t+-+4%29%5E2 is a perfect square and so cannot be negative.
So its minimum value is zero.
So %28t+-+4%29%5E2+=+0
or t = 4
Hence h(t) is maximum when t = 4 seconds.
So maximum height reached is h(4) = 256 ft.

When the object will reach the ground, its height will be zero.
So h%28t%29+=+0+=+16%5E2+-+16%28t+-+4%29%5E2
or 16%5E2+-+16%28t+-+4%29%5E2+=+0
or 4%5E2+-+%28t+-+4%29%5E2+=+0
or t(8-t) = 0
So either t = 0 or 8 - t = 0 i.e. either t = 0 or t = 8
Now, at t = 0, the object was projected up.
So at t = 8 seconds the object must come down to ground.
Thus total time of flight is 8 seconds.

Note: This problem could also have been solved using Maxima & Minima principle of Calculus. But as the asker put it in quadratic section so I have solved this in this method.