SOLUTION: a fualty firworks rocket is launched vertically with an initial velocity of 124ft/s and then falls to the earth unexploded. Its height h in feet after t seconds is given by h(t)=

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Question 289650: a fualty firworks rocket is launched vertically with an initial velocity of 124ft/s and then falls to the earth unexploded. Its height h in feet after t seconds is given by h(t)= -16(Tsquared) + 124T. During which time interval after launch will the height of the rocket exceed 100 feet.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
h%28t%29=+-16t%5E2+%2B+124t
It might help to look at the graph:
graph%28400%2C+400%2C+-2%2C+10%2C+-50%2C+250%2C+-16x%5E2%2B124x%29}
From this we can see that there is a continuous set of times where the rocket is over 100 feet. The rocket will be higher than 100 feet just after it first reaches 100 feet high and it will stay over 100 until the height is 100 feet high for a second time. To find this interval of time we just have to find the times when the rocket is 100 feet:
100+=+-16t%5E2+%2B+124t
Since this is a quadratic equation we will start by getting one side of the equation equal to zero. Adding 16t%5E2 to and subtrcacting 124t from each side we get:
16t%5E2+-+124t+%2B+100+=+0
Next we'll factor:
4%284t%5E2+-+32t+%2B+25%29+=+0
Since the second factor does not factor any further, we'll resort to using the Quadratic Formula:
t+=+%28-%28-32%29+%2B-+sqrt%28%28-32%29%5E2+-+4%284%29%2825%29%29%29%2F2%284%29
t+=+%2832+%2B-+sqrt%281024+-+400%29%29%2F8
t+=+%2832+%2B-+sqrt%28624%29%29%2F8
t+=+%2832+%2B-+sqrt%2816%2A39%29%29%2F8
t+=+%2832+%2B-+sqrt%2816%29%2Asqrt%2839%29%29%2F8
t+=+%2832+%2B-+4sqrt%2839%29%29%2F8
t+=+%284%288+%2B-+sqrt%2839%29%29%29%2F%284%2A2%29
t+=+%28cross%284%29%288+%2B-+sqrt%2839%29%29%29%2F%28cross%284%29%2A2%29
t+=+%288+%2B-+sqrt%2839%29%29%2F2
So the rocket is over 100 feet between
%288+-+sqrt%2839%29%29%2F2 seconds
and
%288+%2B+sqrt%2839%29%29%2F2 seconds.