Question 672600: A soccer ball is kicked in the air the height (h)t of the ball in metres according to the time t in seconds is expresses as a function (h)t = -2t^2+8t. For what interval of the t does the height of the ball increase?
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! There are a couple of ways to "solve" this problem. I'll give you an explanation and solution based on algebra I. Have you studied parabolas yet? If so, you may readily understand this solution, if not I'll expand a little on the steps.
Given
(1) h(t) = -2t^2 + 8t, which we recognize as the equation of a parabola that opens DOWNWARD because the coefficient of t^2 is negative. On flat earth the parabola at t=0 starts at
(2) h(0) = -2*0^2 + 8*0 or
(3) h(0) = 0 meters
Then the ball rises on a parabalic path to a maximum and then "falls" on the same parabolic path to hit the ground where (1) is again equal to zero. Algebraically the start and end of the parabola correspond to the roots of the parabola given by (1) which are obtained by factoring (1) or
(4) h(t) = -2t*(t - 4)
Setting h(t) = 0 we get
(5) -2t*(t-4) = 0 or
(6) t = {0,4}
We know from (3) above that the height is zero at t = 0, now we know that the football hits the ground at t = 4 seconds. It turns out that the soccer ball reaches a maximum height in the "middle" of its trajectory at time, t, equal to one half the time it hits the ground again or
(7) t = 4/2 or
(8) t = 2 seconds
At this time the height is
(9) h(2) = -2*2^2 + 8*2 or
(10) h(2) = -8 + 16 or
(11) h(2) = 8 meters
Now visualize what is occurring. The soccer ball is kicked at time zero at a height of zero, then it rises along a curved parabolic trajectory for two seconds to reach a maximum height of eight meters. Then it begins to "fall" along the parabolic trajectory to hit the ground (height is zero) at time equal to four seconds. This whole trajectory is symmetrical about the time of two seconds.
The answer to the question is that the football rises for two seconds.
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If you a familiar with parabolic equations, you'll recognize the standard form
(12) a*x^2 + b*x + c = 0
You are also that the axis of symmetry is given by
(13) x = -b/(2a)
Comparing your equation (1) to (12), we find that
(14) a = -2
(15) b = 8
(16) c = 0
Puting these values into (13) we get
(17) x = -8/-4 or
(18) x = 2 is the axis of symmetry of your equations.
Since x is equivalent to your t we get t = 2 is the axis of symmetry for your parabola.
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Another way to solve this problem is the use of differential calculus, where we set the derivitive of (1) with respect to t equal to zero and get
(19) -2*2t +8 = 0 or
(20) t = 2
At t=2 the slope (as given by the derivitive) of the trajectory is zero. This means the trajectory hit the apex (peak in this case) at t=2 then start down.
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PS The reason we are on earth for this problem, is that it is the force of gravity that causes the ball to follow a parabolic trajectory.
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