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Homework Help: Kinematics velocity help

  1. Jan 19, 2013 #1
    1. The problem statement, all variables and given/known data

    I am not going to use the exact same problem, as that would be cheating as this is for school, but Instead i will use an example problem:

    A can of pop is tossed directly upwards with an initial velocity of 4 m/s. How high will it go???

    2. Relevant equations

    I know that gravity is -9.8m/s/s so that is my acceleration

    I know the initial velocity is 4m/s

    and I know the final velocity is 0m/s

    3. The attempt at a solution

    What I instinctively do is simply, the initial velocity divided by the maceration which gets me 0.41 but I don't know if that is correct or if that would give me the time it took or the distance it traveled.

    As this is not an actual problem from my homework, only an example problem to help me understand a concept, any help would be appreciated.

  2. jcsd
  3. Jan 19, 2013 #2


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    I assume you meant acceleration. (Maceration is something to do with brewing.) Velocity has dimension LT-1 (distance/time), accn has dimension LT-2, so velocity/accn gives T.
    Easiest way to get the height is by conservation of energy.
  4. Jan 19, 2013 #3

    rude man

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    k.e + p.e. = constant
  5. Jan 19, 2013 #4
    What kinematics formulas do you know that may be relevant to this problem?
  6. Jan 19, 2013 #5
    ΔV= (Vf - Vi) = at

    Vf = 0, Vi = 4, a = g = -9.8

    (0 - 4) = (-9.8)t

    - 4 = -9.8t

    .41 = t
    Last edited: Jan 19, 2013
  7. Jan 19, 2013 #6
    vector wise, if you start somewhere and go up in the air and come back down the resultant will be 0 because you landed in the same spot you started from right?

    consider that a secret hint

    break it down into what you know:
    a= -9.8m/s
    Vf= _____

    can you get 3 of those to solve for a 4th one?
  8. Jan 20, 2013 #7
    Use: [itex]v^2=u^2+2as[/itex]

    where v is final velocity = 0, u is inital velocity = 4m/s, and a is -9.8m/s^2. s is the displacement which is what you are looking for.
    Last edited: Jan 20, 2013
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