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anonymous

  • one year ago

A Punter Kicks A Football . Its Height (h) in meters ,t seconds after the kicks is given by the equation :h=-4.9t^2 + 18.24t +0.8 . The Height Of An Approaching Blockers Hands Is Modeled By Equation : h=-1.43t + 4.26 , using the same time . Can The Blocker Knock Down The Punt ? If So , At What Time Does It Happen ?

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  1. jtvatsim
    • one year ago
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    This one is a little tricky, but I think the plan should be: 1) Set the equations equal to each other (we want to see if the football touches the hands). 2) We will probably need to use the quadratic formula to solve. Do you want to give that a try? I'll help guide you through it.

  2. anonymous
    • one year ago
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    yes

  3. jtvatsim
    • one year ago
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    Alright, let's start with the first step. We will set the equations equal to start: -4.9t^2 + 18.24t + 0.8 = -1.43t + 4.26

  4. jtvatsim
    • one year ago
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    Now, we should probably try to set the equation equal to zero. I'd suggest moving the -1.43t + 4.26 to the other side, like this:

  5. jtvatsim
    • one year ago
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    -4.9t^2 + 19.67t +5.06 = 0 Do you see how I got that?

  6. anonymous
    • one year ago
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    Umm I Think So

  7. anonymous
    • one year ago
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    Adding Correcct? 1.43 + 4.26 TO BOTH SIDES

  8. jtvatsim
    • one year ago
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    Oh wait, lol. Good catch. I was supposed to add the 1.43t, but subtract the 4.26. Sorry about that. :)

  9. jtvatsim
    • one year ago
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    -4.9t^2 + 19.67t - 3.46 = 0 There... that's better. :P

  10. jtvatsim
    • one year ago
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    That should make more sense now. OK so far?

  11. anonymous
    • one year ago
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    YES

  12. jtvatsim
    • one year ago
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    Excellent! Alright, now for the "ugly" part. We have really nasty decimal numbers here, so we should probably just use the quadratic formula. Do you remember the formula or should we look it up?

  13. anonymous
    • one year ago
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    i Remember

  14. jtvatsim
    • one year ago
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    Alright, so I'll trust you and spare you the pain of typing it all out. I will go ahead and plug the numbers into the formula, and we can compare answers after. :)

  15. jtvatsim
    • one year ago
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    What we can definitely say is that: a = -4.9 b = 19.67 and c = -3.46 for when we plug in.

  16. jtvatsim
    • one year ago
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    \[\frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-19.67\pm\sqrt{19.67^2-4(-4.9)(-3.46)}}{2\cdot -4.9}\]

  17. anonymous
    • one year ago
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    Exactly Correct

  18. jtvatsim
    • one year ago
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    Perfect! Now to simplify that horrid looking nightmare. I'll use a calculator now...

  19. jtvatsim
    • one year ago
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    One answer is t = 0.184... the other....

  20. jtvatsim
    • one year ago
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    t = 3.830

  21. anonymous
    • one year ago
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    Yes

  22. jtvatsim
    • one year ago
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    OK, but now what? We have two possibilities either the blocker grabs the ball at t = 0.184 or at t = 3.830... we need a way to figure out which one.

  23. jtvatsim
    • one year ago
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    But we should at least celebrate our progress so far. YAY!!! :)

  24. anonymous
    • one year ago
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    Yay !!!!!!!!!!!!! Lol

  25. jtvatsim
    • one year ago
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    OK, let's do a sanity check with our answers. A useful algebra trick is to plug answers back into the original equations to see if they make sense.

  26. jtvatsim
    • one year ago
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    Let's not use the -4.9t^2 + ... mess, instead I'll check the other equation: -1.43t + 4.26

  27. anonymous
    • one year ago
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    Right Thats Just Makes Us Go To Far.

  28. jtvatsim
    • one year ago
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    OK, plugging in t = 0.184, I see h = 3.99 But t = 3.830, I see h = -1.22

  29. jtvatsim
    • one year ago
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    Well, there's a big problem with the second answer. Do you see it? The blocker's hands are a "negative height," in other words he blocks the ball underground. Unless he is a gopher, I think we can conclude that we should go with the first answer t = 0.184

  30. anonymous
    • one year ago
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    No I Didnt See It , i Just Re-Read The Problem WOW

  31. anonymous
    • one year ago
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    Almost Missed It

  32. jtvatsim
    • one year ago
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    Yep, as I mentioned when we first started, this one was a little sneaky. But hopefully that makes sense overall. Any final questions on this one? :)

  33. anonymous
    • one year ago
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    Yes .

  34. anonymous
    • one year ago
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    Well I Have A Question Mind Helping With Another Onne ?

  35. jtvatsim
    • one year ago
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    Go ahead and close this one and post a new one, so it doesn't get too cluttered. See you there! :)

  36. anonymous
    • one year ago
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    Ok

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