SOLUTION: If a rocket is fired vertically into the air with a speed of 304 feet per second, its height at time t seconds is given by h=-16t2+304t. At what time(s) will the rocket be the

Question 1054873: If a rocket is fired vertically into the air with a speed of
304 feet per second, its height at time t seconds is given by h=-16t2+304t.
At what time(s) will the rocket be the following number of feet above the ground?
a. The rocket will 1440 feet high after?

b. The rocket will 1248 feet high after

c. How long will the rocket be in the air? (Hint: How high is it when it hits the ground?)

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
h=-16t^2+304t
1440=-16t^2+304t
-16t^2+304t-1440=0
divide by -16
t^2-19t+90=0
(t-9)(t-10)=0
at 9 and 10 seconds.
At 9 seconds, -1296+2736=1440. At 10 seconds, -1600+3040=1440.
for 1248
16t^2-304t+1248=0
t^2-19t+78=0
t=(1/2)(19+/- sqrt (361-332); sqrt 29=5.39
t=(1/2)(13.61)=6.8 sec.;12.2 sec is other root.
the vertex is t=-b/2a or -304/-32=9.5 sec
The vertex is the axis of symmetry. The rocket will be in the air 19 seconds.

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