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Math Help - Kinematics problem

  1. #1
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    Post Kinematics problem

    OK here is a kinematic problem which I simply cannot solve... to be honest it has a really easy solution but I just cannot figure it out. (I know I'm missing something)

    A stone is projected vertically upwards with a speed of 14 m/s, from a point O at the top of a mine shaft. Five seconds earlier a mine began to descend the mine shaft from O with a constant speed of 3.5 m/s. Find the depth of the lift (to the nearest metre) at the instant when the stone falls on it. (Neglect air resistance and take g=9.8m/s^2)

    I tried to solve this using velocity time graph and god knows what else...

    another quick question- when the object is goint vertically upwards g= -9.8 right?? and when it is going down would g=+9.8?
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  2. #2
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    Quote Originally Posted by ziggy777 View Post
    OK here is a kinematic problem which I simply cannot solve... to be honest it has a really easy solution but I just cannot figure it out. (I know I'm missing something)

    A stone is projected vertically upwards with a speed of 14 m/s, from a point O at the top of a mine shaft. Five seconds earlier a mine began to descend the mine shaft from O with a constant speed of 3.5 m/s. Find the depth of the lift (to the nearest metre) at the instant when the stone falls on it. (Neglect air resistance and take g=9.8m/s^2)

    I tried to solve this using velocity time graph and god knows what else...

    another quick question- when the object is goint vertically upwards g= -9.8 right?? and when it is going down would g=+9.8?
    There are many approaches. Here is one:

    Start the clock from when the the stone is projected. Then:

    For the lift: x = (5)(3.5) + 3.5 t = 17.5 + 3.5t .... (1)

    For the stone: x = -14t + 4.9t^2 .... (2)

    Solve equations (1) and (2) simultaneously for x.


    Note: I got equation (2) as follows:

    Take downwards as the positive direction. Then:

    a = 9.8 m/s^2
    u = -14 m/s
    t = t
    x = ?

    x = ut + \frac{1}{2} at^2.


    For checking purposes:

    Spoiler:
    x = 32.85 m
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    There are many approaches. Here is one:

    Start the clock from when the the stone is projected. Then:

    For the lift: x = (5)(3.5) + 3.5 t = 17.5 + 3.5t .... (1)

    For the stone: x = -14t + 4.9t^2 .... (2)

    Solve equations (1) and (2) simultaneously for x.


    Note: I got equation (2) as follows:

    Take downwards as the positive direction. Then:

    a = 9.8 m/s^2
    u = -14 m/s
    t = t
    x = ?

    x = ut + \frac{1}{2} at^2.


    For checking purposes:

    Spoiler:
    x = 32.85 m


    hmmmm..... yes that really solves the problem... many thanks
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