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Math Help - Kinematics: The Motion of Particles

  1. #1
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    Kinematics: The Motion of Particles

    At the same time as a stone is dropped from rest from a point A at the top of a 120m high cliff a ball is thrown vertically up with speed 30m/s from a point B which is at the bottom of the cliff and immediately below A. Find the height above B of the point where the ball and the stone collide.

    many thanks!
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  2. #2
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by BabyMilo View Post
    At the same time as a stone is dropped from rest from a point A at the top of a 120m high cliff a ball is thrown vertically up with speed 30m/s from a point B which is at the bottom of the cliff and immediately below A. Find the height above B of the point where the ball and the stone collide.

    many thanks!
    Both of these are constant-acceleration questions.

    You want an expression for height in terms of time for both items. I'd be tempted to try s = ut + \frac {at^2}2 and see what happens.
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  3. #3
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    would it be 30t+0.5*-9.8t^2=0t+0.5*-9.8t^2?
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  4. #4
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by BabyMilo View Post
    would it be 30t+0.5*-9.8t^2=0t+0.5*-9.8t^2?
    Not quite, because one starts at the ground, one at 120m up. So the RHS should be (I think) 120 minus all that.

    That gives you the time at which they hit. Then you just need to plug that back in to get the distance at they reach at that time.
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  5. #5
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    30t-4.9t^2=120+4.9t^2?
    still cant work it out
    can you please show the full working.
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  6. #6
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by BabyMilo View Post
    30t-4.9t^2=120+4.9t^2?
    still cant work it out
    can you please show the full working.
    Okay, you given it a fair shot, let's see ...

    Let u_1, u_2 be the initial velocities of the ball and stone respectively.

    Let s_1, s_2 be the positions above ground of the ball and stone respectively.

    Let a and t be acceleration and time from start respectively.

    Then we have:

    s_1 = 120 + u_1 t - \frac {a t^2} 2

    s_2 = u_2 t - \frac {a t^2} 2

    Then we have u_1 = 0, u_2 = 30. We want s_1 = 120 + s_2:

    30 t - \frac {a t^2} 2 = - \frac {a t^2} 2 + 120

    you should be able to take it from there.
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  7. #7
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    still confess to be honest

    from your equation

    at^2 on both side should cancel each other right?
    which leave us with 120=30t
    therefore t=4

    but when i put this in for x or s = ut+ 0.5*a*t^2
    when
    u=30
    a=-9.81
    t=4
    x therefore = 198

    and yes it works for the other particle
    120+78=198

    but this is not what it's say in my answer on the back.

    the answer should 41.6m

    many thanks!

    sorry for wasting your time!
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  8. #8
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by BabyMilo View Post
    still confess to be honest

    from your equation

    at^2 on both side should cancel each other right?
    which leave us with 120=30t
    therefore t=4

    but when i put this in for x or s = ut+ 0.5*a*t^2
    when
    u=30
    a=-9.81
    t=4
    x therefore = 198

    and yes it works for the other particle
    120+78=198

    but this is not what it's say in my answer on the back.

    the answer should 41.6m

    many thanks!

    sorry for wasting your time!
    a is minus, yeah? You've added.


    You're all right. This stuff is confusing in the early stages. (You're not wasting my time, I'm babysitting a program running on my other computer.)
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  9. #9
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    got it thanks!
    good luck with your program!
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  10. #10
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by BabyMilo View Post
    got it thanks!
    good luck with your program!
    It finished, did what it was supposed to. Successful day.
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