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Physics motion problem

  1. Apr 23, 2009 #1
    1. The problem statement, all variables and given/known data
    A monkey is standing on the top of a cliff 50 metres high and drops a coconut. His friend is standing at the base of the cliff and throws a coconut upward with an initial velocity of 15m/s. At what distance and at what time will the two coconuts collide?

    2. Relevant equations
    v2 = u2 + 2as

    3. The attempt at a solution
    Monkey throwing down:
    v2 = u2 + 2as
    v2 = 2(9.81m/s2)(50m)
    v2= 981.m/s
    v = 31.m

    a = (v2 - v1) / t
    a = (31.m/s - 0m/s) / t
    a = 31.m/s / t
    t = 31.m/s / 9.81m/s2
    t = 3.1s

    Monkey throwing up:
    v2 = u2 + 2as
    0m/s = (15m/s)2 + 2(-9.81m/s2)s
    0m/s = 225m/s - 19.6m/s2s
    -225m/s = -19.6m/s2s
    s = 11.5m

    a = (v2 - v1) / t
    -9.81m/s2 = (0m/s - 15m/s) / t
    -9.81m/s2 = -15m/s / t
    t = -15m/s / -9.81m/s2
    t = 1.5s

    I have the initial and final velocities, the distances, the acceleration, and the times for both coconuts. I'm not sure how to find when the balls actually meet and where they meet. At first, I thought you were supposed to subtract the two times but then I realized that doesn't actually give you the time when the coconuts collided. Thank you in advance for your help.
     
  2. jcsd
  3. Apr 23, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi 5.98e24! Welcome to PF! :smile:
    Nooo … you need the two coconuts to have the same s at the same t, so you need a constant acceleration equation that involves s and t :wink:
     
  4. Apr 23, 2009 #3
    It looks like they collide after the coconut going up starts going down, so:

    My attempt at the solution

    Monkey throwing up coconut 2:
    v2 = u2 + 2as
    0m/s = (15m/s)2 + 2(-9.81m/s2)s
    0m/s = 225m/s - 19.6m/s2s
    -225m/s = -19.6m/s2s
    s = 11.5m

    a = (v2 - v1) / t
    -9.81m/s2 = (0m/s - 15m/s) / t
    -9.81m/s2 = -15m/s / t
    t = -15m/s / -9.81m/s2
    t = 1.5s

    Monkey dropping down coconut 1:
    delta s = vt+.5at^2
    delta s = (0)t + -4.9t^2
    t = 1.5s
    delta = -4.9(1.5)^2
    delta s = -11.025 m, meaning the coconuts have not collided by that point 1.5s
    50 - 11.025 = 38.975 meters
    location of coconut 2 is 11.5 m
    distance between the two at 1.5s = 38.975 - 11.5 = 27.475 m

    Then, I'll need its velocity at that point, so:
    vf^2 = vi^2 + 2as
    vf^2 = 0 + 2*9.8*11.025
    vf = -14.7 m/s

    So, setting up the problem from there:

    coconut 2
    vi = 0
    vf= ?
    t = tcollision
    delta s = x
    a = -9.8

    coconut 1
    vi = vf (see above) = -14.7 m/s
    vf = ?
    delta s = x +27.475
    t = tcollision
    a = -9.8

    Combining the two:
    delta s = vt+.5at^2
    coconut 1
    x = 0(t) + 4.9 t^2
    coconut 2
    x+11.025 = 14.7t + 4.9t^2
    x = 14.7t + 4.9t^2 - 27.475

    Combine: 4.9t^2 = 4.9t^2 + 14.7t - 27.475
    Solve for t, = 1.87 seconds

    .75 + 1.5 = 3.36 seconds

    Plug in t(1.87) and Solve for x, = 17.1 meters
    11.5 - 17.1 meters = - 6, which is impossible

    I dunno; Maybe they collide at 3.1 seconds on the ground.
     
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