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Math Help - Kinematics Question

  1. #1
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    Kinematics Question

    1. A stone is thrown vertically upward from the ground and just reaches the top of a building 100m high. At the same instant that the stone is thrown from the ground a ball is dropped from rest at the top of the building. At what height do the stone and the ball pass one another. (answer: 75m above ground)

    Given:

    Stone
    d_1 = 0m
    a = 9.8m/s^2 [down]
    v_1 = 0m/s Is this velocity actually given? Do I only state a given velocity of 0m/s if "rest" or "stationary" is mentioned?
    d_2 = ?
    v_2 = ?
    \Delta{t} = ?
    \Delta{d} = ?

    Ball
    d_1 = 100m[up]
    a = 9.8m/s^2 [down]
    v_1 = 0m/s
    d_2 = ?
    v_2 = ?
    \Delta{t} = ?
    \Delta{d} = ?

    Required:

    d ball and stone meet = ?

    Analysis:

    Solve for time first.

    Ball
    \Delta{d} = v_1(\Delta{t}) + 0.5a(\Delta{t})^2
    d_2 - d_1 = v_1(\Delta{t}) + 0.5a(\Delta{t})^2

    Stone
    d_2 - d_1 = v_1(\Delta{t}) + 0.5a(\Delta{t})^2

    Determine the height at which they meet.

    Ball
    \Delta{d} = v_1(\Delta{t}) + 0.5a(\Delta{t})^2

    Verify with stone equation

    \Delta{d} = v_1(\Delta{t}) + 0.5a(\Delta{t})^2

    Solve:

    Ball
    d_2 - 100m[up] = v_1(\Delta{t}) + 0.5a(\Delta{t})^2
    d_2 - 100m[up] = 0.5a(\Delta{t})^2
    d_2 = 100m[up] + 0.5a(\Delta{t})^2

    Stone
    d_2 - 0m = v_1(\Delta{t}) + 0.5a(\Delta{t})^2
    d_2 = v_1(\Delta{t}) + 0.5a(\Delta{t})^2

    Sub ball equation into stone equation
    100m[up] + 0.5a(\Delta{t})^2 = v_1(\Delta{t}) + 0.5a(\Delta{t})^2
     100m[up] = v_1(\Delta{t})

    What do I do next? I think I've made an error with the given v_1's, but I'm not sure how.
    Last edited by RogueDemon; April 18th 2010 at 08:23 AM.
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  2. #2
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    any time an object is "dropped" , its initial velocity is zero.


    ball's position equation as a function of time ...

    y = 100 -\frac{1}{2}gt^2

    stone's position equation as a function of time ...

    y = v_0 t - \frac{1}{2}gt^2


    set the position equations equal since they have the same position ...

    100 -\frac{1}{2}gt^2 = v_0 t - \frac{1}{2}gt^2

    100 = v_0 t , where v_0 is the stone's initial velocity upward.


    note the "no-time" equation for constant acceleration ...

    v_f^2 = v_0^2 - 2g(\Delta y)

    at the top of its trajectory, the stone has zero velocity ...

    0 = v_0^2 - 2g(100)

    v_0 = \sqrt{200g}

    sub in for v_0 in the equation 100 = v_0 t ...

    100 = \sqrt{200g} \cdot t

    t = \frac{10}{\sqrt{2g}}

    sub for t in either position equation ...

    y = 100 - \frac{1}{2}g \left(\frac{10}{\sqrt{2g}}\right)^2<br />

    y = 100 - \frac{g}{2} \cdot \frac{100}{2g}

    y = 100 - 25 = 75 \, m
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