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Free fall acceleration (time elapsed, initial & final speed)

  1. Jun 23, 2010 #1
    1. The problem statement, all variables and given/known data
    A rock climber stands on top of a 50-m-high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s apart and observes they cause a single splash. The initial speed of the first stone was 2.0 m/s.

    a) How long after the release of the first stone does the second stone hit the water?
    b) What was the initial speed of the second stone?
    c) What is the speed of each stone as they hit the water?

    2. Relevant equations
    V = Vo + at
    Y - Yo = Vot + .5at2
    v2 = vo2 + 2a(Y - Yo)
    Y - Yo = .5(Vo + V)t

    3. The attempt at a solution
    The problem I am having is that I don't know how to take account the time difference. For instance, you are not just solving for t but have to take into account that one was dropped 1 second after the other. For the first stone, I calculated a time of 2.98 seconds using formula Y - Yo = Vot + .5at2
    Y=-50 m
    Yo=0m
    Vo=-2 m/s
    t=?
    ay=-9.8 m/s^2

    Label Axis: I am assuming the up direction is positive and down direction is negative.

    I can't proceed to parts B and C until I know A. Please Help! Thank you very kindly.
     
  2. jcsd
  3. Jun 23, 2010 #2
    A single splash is seen, hence they hit the water at the same time. Thus, given the initial velocity of the first stone, and the acceleration due to gravity, how long does it take for it to hit the water?

    For b, if it is to cover the same distance in the time it takes a - 1second, what must that stone's initial velocity be?
     
  4. Jun 23, 2010 #3
    So, you solved it :)

    a) Because both stones splash at the same moment - the second stone hit the water 2.98s after the release of the first stone.
     
    Last edited by a moderator: Jun 23, 2010
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