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Math Help - equation of ball

  1. #1
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    equation of ball

    My teacher gave us a brainteaser for extra points... haha but it's been a long time since i used parabolas... guide me to the solution please!! Thank You guys!

    Given equation: h(t)=c-(d-4t)^2

    Problem:
    At time t=0 a ball was thrown upward from an initial height og 6 feet. Until the ball hit the ground, tis height, in feet, after t secons was given by the function h above, in which d and c are positive constants. If the ball reached its maximum height of 106 feet at time t=2.5, what was the heigh in feet of the ball at time t=1?

    Well, I supposed that c=6.. .but that won't make sense when you plug in 2.5 to t.( 6-d^2-100) I am seriously lost.....
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  2. #2
    Junior Member roy_zhang's Avatar
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    Quote Originally Posted by jjkins View Post
    My teacher gave us a brainteaser for extra points... haha but it's been a long time since i used parabolas... guide me to the solution please!! Thank You guys!

    Given equation: h(t)=c-(d-4t)^2

    Problem:
    At time t=0 a ball was thrown upward from an initial height og 6 feet. Until the ball hit the ground, tis height, in feet, after t secons was given by the function h above, in which d and c are positive constants. If the ball reached its maximum height of 106 feet at time t=2.5, what was the heigh in feet of the ball at time t=1?

    Well, I supposed that c=6.. .but that won't make sense when you plug in 2.5 to t.( 6-d^2-100) I am seriously lost.....
    I think the key to solve this question is to find the positive constants c and d. Letís consider the given height function h(t)= c-(d-4t)^2, realize that (d-4t)^2\ge 0, so the maximum height can only be obtained when (d-4t)^2= 0, which means when t=\frac{d}{4} it reaches the maximum height (and the maximum height is h_{\mathtt{max}}=c). Now you found c=106 and \frac{d}{4}=2.5\Rightarrow d=10, then you can answer the question by finding h(1)=106-(10-4)^2

    Roy
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  3. #3
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    Quote Originally Posted by jjkins View Post
    My teacher gave us a brainteaser for extra points... haha but it's been a long time since i used parabolas... guide me to the solution please!! Thank You guys!

    Given equation: h(t)=c-(d-4t)^2

    Problem:
    At time t=0 a ball was thrown upward from an initial height og 6 feet. Until the ball hit the ground, tis height, in feet, after t secons was given by the function h above, in which d and c are positive constants. If the ball reached its maximum height of 106 feet at time t=2.5, what was the heigh in feet of the ball at time t=1?

    Well, I supposed that c=6.. .but that won't make sense when you plug in 2.5 to t.( 6-d^2-100) I am seriously lost.....
    h(t) = c-(d-4t)^2

    We know two points on the parabola in the form (t,h(t)): (0,6), the initial point and (2.5,106), the vertex (maximum)

    Putting the parabola in the form h(t) = a(t-h)^2+k using the point (0,6) for (t,h(t)) and the point (2.5,106) for (h,k) yields:

    6=a(0-2.5)^2+106

    solve for a:

    a=-16

    Now we know our standard equation is h(t)=-16(t-2.5)^2+106

    Replace t with 1 to find the height at 1 second:

    h(1)=-16(1-2.5)^2+106
    h(1)=70

    The height at 1 second is 70 feet.

    I'm not sure why we would need to know the first equation that was given to you. Just using the vertex and initial height at t=0 would suffice to find the height at t=1 second.
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