SOLUTION: Two projectiles are simultaneously launched. The height in meters , h(base 1) t and h(base 2)t of the first and second projectiles as a function of time t in seconds since the laun

Question 754534: Two projectiles are simultaneously launched. The height in meters , h(base 1) t and h(base 2)t of the first and second projectiles as a function of time t in seconds since the launch are given by the rules h(base 1)t = -2t^2+ 20t + 50 and h(base 2) = 2t +66 .
a) How long after their launch will the two projectiles be at the same height?
b) Over what interval of time since the launch is the first projectile higher than the second?
Can you please help me out? Thanks so much in advance:)
Can you also please show the steps it would really help me understand:)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
h(base 1)t = -2t^2+ 20t + 50 and h(base 2) = 2t +66 .
a) How long after their launch will the two projectiles be at the same height?
Solve:
-2t^2 + 20t + 50 = 2t+66
-2t^2 + 18t - 16 = 0
t^2 - 9t + 8 = 0
(t-8)(t-1) = 0
t = 1 second (1st time they are at the same height)
t = 8 seconds (2nd time they are at the same height)
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b) Over what interval of time since the launch is the first projectile higher than the second?
1sec < t < 8 sec
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cheers,
Stan H.
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