SOLUTION: The equation that describes the path of a rocket after it is shot in the air is {{{h=80t-16t^2}}}, where h is the height, in feet, above the ground after t seconds. After how man

Algebra ->  Equations -> SOLUTION: The equation that describes the path of a rocket after it is shot in the air is {{{h=80t-16t^2}}}, where h is the height, in feet, above the ground after t seconds. After how man      Log On


   

Question 553801: The equation that describes the path of a rocket after it is shot in the air is h=80t-16t%5E2, where h is the height, in feet, above the ground after t seconds. After how many seconds will the rocket be at a height of 96 feet?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Set h = 96 and solve the quadratic equation for t.
96+=+80t-16t%5E2 Subtract 96 from both sides and rearrange.
16t%5E2-80t%2B96+=+0 Factor 16 to ease calculations.
16%28t%5E2-5t%2B6%29+=+0 Factor the trinomial.
t%5E2-5t%2B6+=+%28t-2%29%28t-3%29
%28t-2%29%28t-3%29+=+0 Apply the zero product rule:
t-2+=+0 or t-3+=+0 so:
t+=+2 or t+=+3
You have two solutions for this problem which was to be expected because quadratics have two solutions.
At t = 2 seconds, the rocket reaches a height of 96 feet on the way up.
At t = 3 seconds, it will pass the 96-foot level on the way down.