Question 211035: Using the Library, web resources, and/or other materials, find a real-life application of a quadratic function. State the application, give the equation of the quadratic function, and state what the x and y in the application represent. Choose at least two values of x to input into your function and find the corresponding y for each. State, in words, what each x and y means in terms of your real-life application. Please see the following example. Do not use any version of this example in your own post. You may use other variables besides x and y, such as t and S depicted in the following example, but you may not use that example. Be sure to reference all sources using APA style.
Typing hint: To type x-squared, use x^2. Do not use special graphs or symbols because they will not appear when pasted to the Discussion Board.
When thrown into the air from the top of a 50 ft building, a ball’s height, S, at time t can be found by S(t) = -16t^2 + 32t + 50. When t = 1, s = -16(1)^2 + 32(1) + 50 = 66. This implies that after 1 second, the height of the ball is 66 feet. When t = 2, s = -16(2)^2 + 32(2) + 50 = 50. This implies that after 2 seconds, the height of the ball is 50 feet.
I used this. I wanted to know if I was correct.
Real life quadratic function application:
The formula s=-16t^2+vOt+sO
sO=initial height
vO=velocity in feet per second
t=time
The initial height of the mountain where you are standing is 300 feet.
I launch a rock in the air at 52 feet per second.
I want to find out how high it will be after 3 seconds.
Now let’s do the formula.
s=-16t^2+vOt+sO
=-16(3) ^2+52(3) +300
=-16(9) +156+300
=-144+465
s=312
So at 3 seconds, it's 312
This accurately describes the rock's height at any given time from the time of launch to the time.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Good work.
You will find other examples by using Google to search for quadratic
applications.
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Also, the formula does not give the height at "all times": only for the
times the ball is in the air or the instant it reaches the ground.
After that the answers will be negative and therefore not applicable.
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Cheers,
Stan H.
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