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Diver Speed Kinematics motion

  1. Dec 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Two divers jumped from a 3m platform. One jumps upward at 1.8ms^-1 and the second steps off the platform as the first passes it on the way down.

    a) What are their speeds as they hit the water?
    b) Which hits the water first and by how much?

    2. Relevant equations



    3. The attempt at a solution


    a) What are their speeds as they hit the water?

    Diver 1:
    Calculate distance diver 1 travels as he jumps 1.8ms^-1 upwards off the platform.

    vf^2 - vi^2 = 2g(yf - yi)
    0 - (18ms^-1)^2 = 2(-9.8ms^-2)(yf - yi)
    yf - yi = 0.165m
    3 m + 0.165m = 3.165m
    vf^2 - vi^2 = 2g(yf - yi)
    vf = 7.8ms^-1

    diver 1 hits water at 7.8ms^-1

    Diver 2:
    vf^2 - vi^2 = 2g(yf - yi)
    vf^2 - 0 = 2(-9.8ms^-2)(-3m)
    vf = 7.6ms^-1

    Diver 2 hits water at 7.6ms^-1


    b) Which hits the water first and by how much?

    Diver 1:

    At the instant at which diver 1 is 3.165m off the ground, vi = 0
    -3.165m = 0.5(0ms^-1 + (-7.8ms^-1))t
    t = 0.8s

    Diver 2:

    -3 = 0.5(0ms^-1 + (-7.6ms^-1))t
    t = 0.78s

    My answers are wrong according to the book:
    a) -7.67ms^-1
    b) 0.162s
     
    Last edited: Dec 13, 2013
  2. jcsd
  3. Dec 13, 2013 #2

    Curious3141

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    Homework Helper

    For the first part, I would check what value of g the textbook expects you to use, and whether you had round-off errors in intermediate steps. g = -9.81m/s^2 gives the expected answer to 3 s.f.

    Your working in the first part looks OK, but I would be careful of your sign convention. Define up to be positive, and down to be negative. That may explain why the answer key has a negative sign (but they mentioned speed, which shouldn't have a sign - if they'd said velocity, they'd be correct).

    Note that you don't actually have to calculate the point at which Diver 1 comes to momentary rest. Just use initial velocity = +1.8m/s and g = -9.81m/s^2 to get the final velocity in one step. Remember that the s term (or y term, as you're using it) is actually displacement and not total distance. Minimise the number of steps you take to get your answers, and there will be less chance of error.

    As for the second part, I'm assuming you're using ##s = \frac{(u+v)t}{2}## (state the equation you're using - always!).

    For diver 1, you seem to be working out the time taken to reach the water *from the time he is at maximum height and is momentarily at rest*. Is this really the time you want for diver 1? Remember that diver 2 steps off the platform only as diver 1 passes him.
     
  4. Dec 13, 2013 #3
    I'm still getting -7.88ms^-1 after rounding off to 3SF
    If the answer states -7.67ms^-1 then something is wrong
     
  5. Dec 13, 2013 #4

    Curious3141

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    Homework Helper

    The question asks for speeds, plural. Diver 1's speed is 7.88m/s downward (so velocity = -7.88m/s). I'm getting that too, so I think you're right.

    Diver 2's speed is 7.67m/s downward (v = -7.67m/s). Are you getting this?
     
  6. Dec 13, 2013 #5
    I am.

    I'll work on part (b) now
     
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