Question 1169617: H= -5t^2+20t models the height of a ball that is kicked into the air, where H is the ball's height in meters and t is time in seconds after being kicked.
a) How long is the ball in the air for?
b) When will the ball reach its maximum height?
c) What is the maximum height reached by the ball?
d) At what time(s) will the ball be 15 meters in the air?
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! The height of the ball as a function of the time is represented by a parabola.
The vertex of the parabola with equation y = ax^2 + bx + c, is at x = -b/2a.
Thus, the time to reach maximum height = -20/-10 = 2 s. The ball is in the air for
twice that, or t = 4 s.
The maximum height is given by H(2) = -5(2)^2 + 20(2) = -20 + 40 = 20 m
15 = -5t^2 + 20t -> -5(t^2 - 4t + 3) = 0. This factors as (t-3)(t-1) = 0.
Thus, there are two times when the ball is at 15 m, t = 1s and t = 3s.
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