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Numbers gone wrong, free fall

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data

    a stone is dropped into th water fro a bridge 44 m above the water. another stone is thrown vertically 1.0s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?

    2. Relevant equations

    x = x0 + vx0t + 1/2 ax(t)
    vx = vxo + axt
    x = x0 + 1/2(vx+vx0)(t)

    3. The attempt at a solution

    it seemed like a pretty simple question, I calculated how long it took for a stone to go from 44 m above the water to 0 m in free fall ( 0 = 44 - 4.9(t) ) then I subtracted 1 second from that time and plugged it into 0 = 44 + vx(8) - 4.9(8) to see the initial velocity of the second stone..

    then something strange happened.. I tried to plot a speed v.s. time graph and so I wanted to find the velocity of the stone at a point in time.. so I plugged vx = -4.9(9) and I get some pretty big number.. -88.. then I tried plugging it into x = 1/2(-88)(9) and I get a humongous number.. what's going on?

  2. jcsd
  3. Oct 10, 2009 #2
    Hi, firstly your equation is wrong x = x0 + vx0t + 1/2 ax(t). You see a.t will give you the unit of m/s however v.t gives you m which is distance so how can you add speed with distance? Secondly x0 is the initial displacement from a point a reference and x is your final displacement. Try again and i believe your approach is correct :biggrin:
  4. Oct 10, 2009 #3
    I'm not sure :S, that equation should work shouldn't it? I've used it before.. Also, my x0 is 44 because I'm taking that as the initial distance and so my acceleration is negative because it will go from 44m to 0m

  5. Oct 10, 2009 #4
    Yeap you are correct in the displacement part :tongue: however as i mentioned if you were to look at the equation you wrote, the units on the left and right won't add up do they? I believe the equation is x = x0 + vt + 0.5at^2
  6. Oct 10, 2009 #5
    the equation is x = x0 + vx0t (initial velocity) + 0.5at^2 and it should work, as it comes from x = x0 + 1/2(v + vx0 )t which is just the addition of distances..

  7. Oct 10, 2009 #6
    can someone help please..
  8. Oct 10, 2009 #7
    What seem to be your problem? 0 = 44 - 4.9(t) should be 0 = 44 - 4.9(t2)
  9. Oct 10, 2009 #8

    AH right.. I forgot that it was at^2

    thanks a lot!!
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