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Angel J

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An object is thrown upward into the air with an initial velocity of 128 feet per second. The formula h(t) = 128t - 16t^2 gives its height above the ground after t seconds. What is the height after 2 seconds? What is the maximum height reached? For how many seconds will the object be in the air?

How do I go about solving this with quadratic equations (which I know is given)?
 
1- Substitute 2 instead of t to get height. h= 128*2-16*4= 192 unit.. is it feet?

2- max hight is when velocity is 0, we take the first derivative of the eqation and set it equal to 0,
dh/dt= 128- 32t=0, therefore, t= 128/32= 4 seconds when the object reaches the max hight. Therefore, using the first equation and substituting 4 instead of t, mas h= 128*4- 16*16= 256 feet

3- The object will take 4 seconds to reach max hight, therefore, it will take another 4 seconds to come back to the ground. total time is 4+4= 8 seconds
 
it would scientifically reduce its proportion by .369, then reproduct itself to the higher decimal point of .400 this is a long process h(t)- 125 at the end of the equation you it will be.. 300.7.
 
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