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

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) (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 (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=64 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 3, 11. Here's your graph:

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) (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.

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