SOLUTION: A rocket launched from ground level with an initial velocity of 224 ft/s. When will the rocket reach a height of 528 ft? I know that h=vt-16t^2 will be used. So what happens after

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Question 256232: A rocket launched from ground level with an initial velocity of 224 ft/s. When will the rocket reach a height of 528 ft? I know that h=vt-16t^2 will be used. So what happens after I plug it all in? 528=224t-16t^2?
Found 2 solutions by Greenfinch, Alan3354:
Answer by Greenfinch(383) About Me  (Show Source):
You can put this solution on YOUR website!
It is simpler dividing through by 16
33 = 14t -t^2 so
t^2 - 14t + 33 = 0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation at%5E2%2Bbt%2Bc=0 (in our case -1t%5E2%2B14t%2B-33+=+0) has the following solutons:

t%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2814%29%5E2-4%2A-1%2A-33=64.

Discriminant d=64 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-14%2B-sqrt%28+64+%29%29%2F2%5Ca.

t%5B1%5D+=+%28-%2814%29%2Bsqrt%28+64+%29%29%2F2%5C-1+=+3
t%5B2%5D+=+%28-%2814%29-sqrt%28+64+%29%29%2F2%5C-1+=+11

Quadratic expression -1t%5E2%2B14t%2B-33 can be factored:
-1t%5E2%2B14t%2B-33+=+-1%28t-3%29%2A%28t-11%29
Again, the answer is: 3, 11. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-1%2Ax%5E2%2B14%2Ax%2B-33+%29

This gives t = 3 and t = 11. So it will pass 528 feet after three seconds, carry on until it runs out of speed after 7 seconds when it starts to drop until it passes 528 feet on the way down after 11 seconds hits the ground after 14 seconds

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A rocket has a motor and accelerates going up. This equation is for a ballistic object that is launched and is then subject to only the force of gravity.