SOLUTION: 1. (20 points) In 1981, Vadimir Poliakov (USSR) set a world record of 19ft ¾ in for the pole vault (www.polevault.com). To reach that height, Poliakov obtained a speed of approxim

Algebra ->  Graphs -> SOLUTION: 1. (20 points) In 1981, Vadimir Poliakov (USSR) set a world record of 19ft ¾ in for the pole vault (www.polevault.com). To reach that height, Poliakov obtained a speed of approxim      Log On


   



Question 277866: 1. (20 points) In 1981, Vadimir Poliakov (USSR) set a world record of 19ft ¾ in for the pole vault (www.polevault.com). To reach that height, Poliakov obtained a speed of approximately 36 feet per second on the runway. The formula h = -16t2 + 36t gives his height t seconds after leaving the ground.
a. Does the graph of this equation open up or down? How did you determine this?
b. Describe what happens to his height as time passes, starting at t=0?
c. Use the method of your choice to find how long he was in the air.
d. Will his height peak or be at a low during the middle of the time he is in the air? How do you know?
e. After how many seconds will the peak or low occur?
f. How high will he be when the peak or low occurs?
g. What is the point of the vertex? How does this number relate to your answers in parts e and f?
h. How many solutions are there to the equation 0 = -16t2 + 36t? How do you know?
i. To use the quadratic formula to solve an equation like this, we need three values, represented in your text by a, b and c. In the equation in part h, what are a, b & c?
j. What do the solutions to 0 = -16t2 + 36t represent? Is there a solution that does not make sense? If so, in what ways does the solution not make sense?

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
In 1981, Vadimir Poliakov (USSR) set a world record of 19ft ¾ in for the pole vault (www.polevault.com). To reach that height, Poliakov obtained a speed of approximately 36 feet per second on the runway. The formula h = -16t^2 + 36t gives his height t seconds after leaving the ground.
a. Does the graph of this equation open up or down? How did you determine this?
Ans: Down because of the -16t^2
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b. Describe what happens to his height as time passes, starting at t=0?
Ans: h(0) = 0 means the height is zero when the time is zero.
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c. Use the method of your choice to find how long he was in the air.
Ans: Solve h=0 for two values of "t".
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d. Will his height peak or be at a low during the middle of the time he is in the air? How do you know?
Ans: peak because the parabola opens down and therefore has a peak at the vertex.
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e. After how many seconds will the peak or low occur?
Ans: Find the t-value of the peak at t = -b/2a
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f. How high will he be when the peak or low occurs?
Ans: Find h at that t-value.
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g. What is the point of the vertex? How does this number relate to your answers in parts e and f?
Ans: vertex has time of peak and height of peak.
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h. How many solutions are there to the equation 0 = -16t2 + 36t? How do you know?
Ans: It is a quadratic
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i. To use the quadratic formula to solve an equation like this, we need three values, represented in your text by a, b and c. In the equation in part h, what are a, b & c?
Ans a = -16 ; b = 36 ; c = 0
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j. What do the solutions to 0 = -16t2 + 36t represent? Is there a solution that does not make sense? If so, in what ways does the solution not make sense?
Ans: They both tell the time when the vaulter is on the ground.
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Cheers,
Stan H.