Question 186911: A projectile is thrown upward so that its distance,in feet, above the ground after t seconds is h=-12t^2+456t. What is its maximum height?
Answer by Jose Angel(1)   (Show Source):
You can put this solution on YOUR website! If you want to calculate the maximum height and you have the function h(t), you must find its derivative, and in this case we get h'(t)=-24t+456. Then we solve the equation
-24t+456=0 to find the maximum and the minimum. The solution is t=19s.
If we want to proof that it is the time when the height is maximum, we calculate the second derivative, that is, h''(t)=-24, that is negative, so we have in t=19s a maximum (if the number is positive we would have found a minimum).
So when t=19s we get the maximum, that is equal to h(19)=-12(19)^2+456*19=4332m.
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