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Math Help - Physics: Motion problem

  1. #1
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    Physics: Motion problem

    No Calculus allowed. I need some help. I'm really stuck...

    Thank You.


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  2. #2
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    no calculus, huh?

    since \Delta y = 0

    0 = v_{yo}t - \frac{1}{2}gt^2

    0 = t\left(v_{yo} - \frac{1}{2}gt\right)

    v_{yo} = \frac{1}{2}gt

    t = \frac{2v_{yo}}{g} = \frac{2v_o\sin{\theta}}{g}

    \Delta x = v_x \cdot t

    \Delta x = v_o\cos{\theta} \cdot \frac{2v_o\sin{\theta}}{g}

    \Delta x = \frac{v_o^2 \cdot 2\sin{\theta}\cos{\theta}}{g}

    \Delta x = \frac{v_o^2 \sin(2\theta)}{g}

    max value for \Delta x occurs when \sin(2\theta) = 1

    \Delta x_{max} = \frac{v_o^2}{g}


    min ball speed will occur when it is at the peak of its trajectory ... when v_y = 0 ... min speed would therefore be ... ?


    max height occurs at t = \frac{v_{yo}}{g}
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  3. #3
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    Quote Originally Posted by skeeter View Post
    no calculus, huh?

    since \Delta y = 0

    0 = v_{yo}t - \frac{1}{2}gt^2

    0 = t\left(v_{yo} - \frac{1}{2}gt\right)

    v_{yo} = \frac{1}{2}gt

    t = \frac{2v_{yo}}{g} = \frac{2v_o\sin{\theta}}{g}

    \Delta x = v_x \cdot t

    \Delta x = v_o\cos{\theta} \cdot \frac{2v_o\sin{\theta}}{g}

    \Delta x = \frac{v_o^2 \cdot 2\sin{\theta}\cos{\theta}}{g}

    \Delta x = \frac{v_o^2 \sin(2\theta)}{g}

    max value for \Delta x occurs when \sin(2\theta) = 1

    \Delta x_{max} = \frac{v_o^2}{g}


    min ball speed will occur when it is at the peak of its trajectory ... when v_y = 0 ... min speed would therefore be ... ?


    max height occurs at t = \frac{v_{yo}}{g}

    How did you figure all of these side details out like delta y = 0?


    How do you determine which parts of the problem belong to X and which ones belong to y ?

    For the last part can you at least show me the equations that I would use?

    Thanks,
    qbkr21
    Last edited by qbkr21; January 31st 2009 at 02:15 PM.
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  4. #4
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    How did you figure all of these side details out like delta y = 0?

    if the projectile starts on the ground and ends on the ground, what is \Delta y?


    How do you determine which parts of the problem belong to X and which ones belong to y ?

    I don't understand the question ... what parts do you mean?

    For the last part can you at least show me the equations that I would use?

    I gave you the equation for time to the top of the projectile's trajectory ... you need to calculate \Delta y for that time.

    \Delta y = v_{yo}t - \frac{1}{2}gt^2


    you could also ignore the time element and use this equation to calculate \Delta y ...

    v_{yf}^2 = v_{yo}^2 - 2g(\Delta y)

    at the top of its trajectory, v_{yf} = 0
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  5. #5
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    RE:

    Question:

    b) What is the minimum speed of the ball during this hole-in-one shot?

    The book says the answer is 21.2 m/s

    I set up my equation as follows...

    (V_y)^2=(V_xy)^2-2g*delta(y)
    0=V_xy)^2-1799.28
    x^2=1799.28
    x=42.4

    ***Where did I go wrong?
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  6. #6
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    at the top of its trajectory, the projectile's y-component of velocity is zero ... so, it only has an x-component of velocity which is constant since there is no acceleration in the x-direction.

    v_x = v_o\cos{\theta}
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