Question 1200642: A projectile is shot straight up from a height of 6m with an initial velocity of 80m/s. Its height in meters above the ground after t seconds is given by the equation h=6+80t-5t^2. After how many seconds does the projectile reach its maximum height, and what is this height
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A projectile is shot straight up from a height of 6m with an initial velocity of 80m/s. Its height in meters above the ground after t seconds is given by the equation h=6+80t-5t^2. After how many seconds does the projectile reach its maximum height, and what is this height
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h(t) = 6+80t-5t^2
The max of the parabola is at t = -b/2a
--> t = -80/-10 = 8 seconds
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h(8) = 6 + 80*8 - 5*64 = 326 meters
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Or,
h(t) = 6+80t-5t^2
h'(t) = 80 - 10t = 0
t = 8
etc
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