F.4 Quadratic Functions
2009-11-20 4:35 am
An object is projected vertically upward to the sky. After t seconds, the height of the object is h m above the ground and h = 200 - 5t^2. Find the maximum height of the object above the ground.
回答 (1)
2009-11-20 4:41 am
✔ 最佳答案
I believe there is a mistake in the question and the equation should be as follows:
h = 200t - 5t^2
h = -5(t^2 - 40t)
h = -5(t^2 - 40t + 400 - 400)
h = -5(t - 20)^2 + 2000
The maximum height is reached at t = 20 and h = 2000 m
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