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Can someone help explain this?

  1. Sep 7, 2006 #1
    Hi there, I am in college algebra. I haven't done any of this in 3 years and am a bit rusty, I am very far behind and don't understand how to do this! Can someone please not just give me answers but help explain how to do a problem for me? The problem is this...

    An object is propelled vertically upward with an initial velocity of 20 meters per second. The distance s(in meters) of the object from the ground after t seconds is s=4.9t^2+20t

    a) When will the object be 15 meters above the ground?

    b) when will it strike the ground?

    c) Will the object reach a height of 100 meters.

    I am guessing that with the a) you would make s=15 but as for what to do from there I have no idea. Same for be except making s=0?? And as far as c I don't have a clue. How could you answer c without knowing the rate that the object slows down to know if it would ever make it that high? I just don't get it!
     
  2. jcsd
  3. Sep 7, 2006 #2
    the equation you have is slightly wrong... it should be s=-4.9t^2+20t and yes it would be a good idea to set s=15 m and solve for t (remember the quadratic equation?) same for b. as for c set it to 100 and see if you can get a real number (watch out for square roots of negative numers)
     
  4. Sep 7, 2006 #3
    sorry about the misprint! ok so

    -4.9t^2 + 20t = 15 so then do I make it all = to 0. So subtact the 15 from both sides? Or do I divide everything by -4.9 to get the coefficient to the square be 1?
     
  5. Sep 7, 2006 #4
    start by subtracting 15 from both sides... the quadratic equation allows you to have an 'a' values for ax^2+bx+c, it is just usually equal to 1 in easier problems.
     
  6. Sep 7, 2006 #5
    ok so this is what I have done so far...

    -15 + 20t - 4.9t^2=0

    divided everything by -4.9

    +15/4.9 - 20t/4.9 + t^2 = 0

    subtract 15/4.9 from both sides to get -20t/4.9 + t^2 = -15/4.9

    Am I on the right tract of way off? I am not sure what to do now. Do I add the 20t/4.9 to both sides to get the t^2 by itself. And then divde by t on both sides again to get t alone?
     
  7. Sep 7, 2006 #6
    remember the quadratic equation? maybe you forgot so here it is:
    [tex]x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
    you already have your problem set up in the form of [tex]ax^2+bx+c=0[/tex]
     
  8. Sep 7, 2006 #7
    oh boy...I think I am just doomed. lol I don't remember ANYTHING
     
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