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Homework Help: How far apart are the two rocks?

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data
    From the top of a 40m high cliff, two rocks are simultaneously thrown. One is thrown upwards at 20m/s, the other is thrown downwards at 8.0m/s. How far apart are the two rocks after 1.5 seconds?

    2. Relevant equations
    vf = vi + at
    d = Vavg * t
    d = vit + 1/2 at^2
    vf^2 = vi^2 + 2ad

    3. The attempt at a solution
    I've enclosed the above equations, but i've not established which ones' are going to be helpful in arriving at the answer yet. I'm trying find the solution now, but wanted to post it up here to check my solution with whomever comes up with the right answer. Thanks! :-)
  2. jcsd
  3. Sep 27, 2009 #2


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    Homework Helper

    Use d = vit + 1/2 at^2 for each rock. The formula tells you the height of each rock is at any time you specify.
  4. Sep 27, 2009 #3
    I figured that the height for the rock above ground thrown upwards was 60 m. It takes 3.5 seconds for the rock to fall from the top to the ground. 2.0 seconds for the rock to reach the top, and total flight time of 5.5 seconds.
    I still can't figure this one out...help more please?
  5. Sep 27, 2009 #4
    A better approach here would be to use relative kinematics.

    What is the equation of motion of one rock relative to the other?
    Note that there is no relative acceleration between the two, since they're both being accelerated at the same rate (g)

    [tex]\vec x_{12}= \vec x_1 - \vec x_2[/tex] (The displacement of object 1 relative to object 2)

    [tex]\vec v_{12}= \vec v_1 - \vec v_2[/tex]

    [tex]\vec a_{12}= \vec a_1 - \vec a_2[/tex]
  6. Sep 27, 2009 #5
    Thank you, but unfortunately i have to use the formulas we've been working with until i get into relative kinematics. Is there a way you can show me with the formula/s above?
  7. Sep 27, 2009 #6
    Well, why not develop them? :)

    Start by defining the relative distance, [tex]\vec x_{12}[/tex]
    And then differentiate with respect to time to get the relative velocity and acceleration equations. :3

    But if that's not something you're supposed to do, you could always brute-force it.

    Write out the motion equations for each of the rocks, set [tex]t=1.5 sec[/tex] and look at the difference between the two.

    That looks like an awful waste of time, though.
  8. Sep 27, 2009 #7
    well..thanks for trying to help, but i guess i'm not able to even write out the motion equations...because i don't know how or what equations to plug that time into and how i can determine the distance between the two rocks...anyway, thanks for your suggestions.
  9. Sep 27, 2009 #8
    tried plugging in time 1.5s into d=vit+1/2at^2 for vi=8m/s and t=1.5 and got 23m with +9.80m/s^2.
    Then tried same thing except plugged in 20m/s and got 19m with -9.80m/s^2. I know the answer is the rocks are 42m apart at 1.5 seconds, but i still don't see how they arrived at that. Anyone help me out here please?
  10. Sep 27, 2009 #9
    Remember that [tex]\vec v_i[/tex] and [tex]\vec a[/tex] are vector quantities. That means that their direction is important. Now look at your equations, did you account for their directions or not?
  11. Sep 27, 2009 #10
    Okay, i see what you mean, but we're first year intro physics and all i can see from the original question is two directions...up and down. So i'm not sure what i am missing here...then again...i'm new at physics...please help?
  12. Sep 27, 2009 #11
    Exactly, and did you account for that in your equations? Make a diagram of both the rocks, with their initial displacements, velocities and accelerations. I suggest you choose the edge of the cliff as the origin for your calculations. That would make their initial displacements both zero.
    Now take a good look at the signs of their displacement after 1.5 seconds. With regards to each of them, are the below or above the edge of the cliff? What does that tell you about the distance between the two?
  13. Sep 27, 2009 #12


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    Rock 1: d = vit + 1/2 at^2 = 20*1.5 - 4.905*1.5^2 = 18.96 m
    Rock 2: similar

    When you have both heights at time 1.5, take the difference!
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