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Question 99594: If a soccer ball is kicked straight up from the ground with an initial velocity of 32 feet per second, then it's height above the earth in feet is given by s(t)=-16t^2+32T where t is time in seconds. Graph this parabola for 0 is less than or equal to t less than or equal to 2. What is the maximum height reached by the ball?
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! You have a generous teacher who selected the value of the initial upward velocity of the ball
such that it will help simplify the problem.
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The equation you were given for the height is:
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where t is the time that the ball is in flight.
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Think about the flight of the ball. It starts at a height of zero (ground level). At time
t = 0 it gets kicked and it rises up into the air. But gravity is acting on it. As the ball
rises, gravity slows it down until at some height the ball stops going up and starts to
fall back down toward the ground. It continues to fall until it hits the ground and at that
time its height is again zero.
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So let's set the height s equal to zero and solve the equation to see at what values of time
the height is zero:
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But notice that both the terms on the right side contain -16 as a factor. Therefore,
we can simplify the problem by dividing both sides (all terms) by -16 and the equation
reduces to:
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Now notice that both terms on the right side contain t. Therefore we can factor t and the
equation becomes:
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Finally notice that if either t = 0 or t - 2 = 0 the right side will involve a multiplication
by zero and will therefore will equal the left side. What we are doing is solving
for the values of t which make the height s equal to zero. We already knew that the height
of the ball would be zero when t = 0, because that was the height at the instant the ball
was kicked. When we solve t - 2 = 0 we get an answer of t = 2. That's what we didn't know.
At 2 seconds the ball will be back on the ground at zero height. Ignoring wind resistance
the time of the flight of the ball upward will equal the time it takes to fall back to
Earth. Since the total flight takes 2 seconds, the ball spends 1 second going up and
1 second falling back to ground. Therefore, its maximum height will occur 1 second after
you kick it. To find the height at 1 second, return to the equation:
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and substitute 1 second for t:
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The maximum height of the ball is 16 feet and it occurs 1 second after the ball is kicked.
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You can find other values for your graph. You know that at t = 0, s = 0 and at t = 2, s = 0, and
at t = 1, s = 16. So the points (0,0), (2,0), and (1, 16) are on the graph. You need to try some
fractional values of t. For example, set t = 1/2 and use the equation:
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to find the corresponding value of s.
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Another thing that will help you is to recognize that if you draw a vertical line through
the point t = 1, you will see that the horizontal distance from the graph on one side to
this vertical line will be the same as the same horizontal distance from the vertical line
to the graph on the other side of the line.
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Your graph should look like this:
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This graph may help you to see that if you draw a vertical line through the point 1 on the
horizontal axis, and for any height you draw a horizontal line through the graph, the
distance from the graph to the vertical line is the same on both sides of the vertical
line. This is referred to a symmetry. The graph looks the same on both sides of that vertical
line
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Hope this helps you with your problem.
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