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Free fall physics velocity

  1. Sep 14, 2014 #1
    1. The problem statement, all variables and given/known data
    While standing on a bridge 18.3 m above the ground, you drop a stone from rest. When the stone has fallen 2.50 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction.


    2. Relevant equations
    Free fall problem


    3. The attempt at a solution

    1) stone 1:
    V0 = 0
    a= -9.8 m/s^2
    y= 18.3m - 2.50m = 15.8m

    Stone 2:
    a= -9.8 m/s^2
    y= -???
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 14, 2014 #2
    This is my take on it, but I encourage corrections where there may be mistakes:

    Stone 1: Δx=15.8 m
    V0= 0 m/s
    a= +9.8 m/s2

    Use Formula: Δx=V0t + 1/2at2
    15.8=4.9t2
    3.2=t2
    1.8 s=t

    Stone 2: Δx= 18.3 m
    V0= ?
    a= +9.8 m/s2

    Use Formula again, substitute t= 1.8 to get 1.3 m/s for V0
     
  4. Sep 14, 2014 #3
    why is a=+9.8m/s^2? I thought acceleration would be Negative because you are going downward
     
  5. Sep 14, 2014 #4
  6. Sep 14, 2014 #5
    so from where you left off

    X= Vo t + 1/2 a t

    18.3m = Vo(1.8s) + 1/2(9.8m/s^2) (1.8)^2

    Vo = -1.35 m/s
     
  7. Sep 14, 2014 #6
    Close. Just change the sign to positive so your answer should be 1.35 m/s.
     
  8. Sep 14, 2014 #7

    Simon Bridge

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    The sign convention is arbitrary. You can use any convention as long as you are consistent. When you start a problem, you should state what convention you are using.

    For many problems it is convenient to choose the positive direction to be "downwards", in which case you write "+y = down", which means that gravity is positive, so: ag=+9.8m/s/s = g

    This also means that a positive initial velocity will be downwards, and positive distances are below the starting height.

    You chose to make "+y = up" so that gravity is negative, so: ag=-9.8m/s/s = -g
    This probably makes more intuitive sense, but can make mistakes easier.

    Basically the acceleration vector points in the same direction as the net force.
     
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