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  1. #1
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    attempted this physics problem.

    A 47.0 g golf ball is driven from the tee with an initial speed of 48.0 m/s and rises to a height of 24.8 m.
    (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
    J

    (b) What is its speed when it is 10.0 m below its highest point?
    m/s

    not sure why b won't be given by vf = vo^2 + 2*g*(yo - yf)
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  2. #2
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    Quote Originally Posted by rcmango View Post
    A 47.0 g golf ball is driven from the tee with an initial speed of 48.0 m/s and rises to a height of 24.8 m.
    (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
    J

    (b) What is its speed when it is 10.0 m below its highest point?
    m/s

    not sure why b won't be given by vf = vo^2 + 2*g*(yo - yf)
    (a)

    $\displaystyle (v_{f})^2=(v_{i})^2+2ad$

    $\displaystyle 0=48.0^2+2(-9.81)d$

    $\displaystyle 0=2304-19.62d$

    $\displaystyle \frac{2304}{19.62}=d$

    $\displaystyle 117.43=d$

    $\displaystyle K.E.=\frac{1}{2}mv^2$

    $\displaystyle 0=\frac{1}{2}(0.047)(0)^2$

    (b)

    $\displaystyle P.E.=K.E.$

    $\displaystyle mgh=\frac{1}{2}mv^2$

    $\displaystyle (0.047)(9.81)(117.43-10)=\frac{1}{2}(0.047)v^2$

    $\displaystyle 49.53=\frac{1}{2}(0.047)v^2$

    $\displaystyle \frac{99.07}{0.047}=v^2$

    $\displaystyle \sqrt{2107.78}=v$

    $\displaystyle 45.91=v$
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  3. #3
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    Quote Originally Posted by rcmango View Post
    A 47.0 g golf ball is driven from the tee with an initial speed of 48.0 m/s and rises to a height of 24.8 m.
    (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
    J
    I think hummeth assumed that the ball was hit straight upward. This is clearly not true as an object projected directly upward with a speed of 48 m/s will attain a maximum height of 117.6 m, not 24.8 m.

    Let upward be the positive y direction, and let the zero point for the gravitational potential energy be at the level of the tee. There are no losses due to friction so we know that
    $\displaystyle \Delta E = 0$

    $\displaystyle \Delta KE + \Delta PE = 0$

    $\displaystyle \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 + mgh - mgh_0 = 0$

    Now, the ball started with a speed $\displaystyle v_0$ equal to 48.0 m/s and at a height $\displaystyle h_0$ equal to 0 m. The maximum height h is given as 24.8 m, so the KE at that point is:
    $\displaystyle \frac{1}{2}mv^2 - \frac{1}{2}(0.047)(48.0)^2 + (0.047)(9.8)(24.8) = 0$

    Thus
    $\displaystyle \frac{1}{2}mv^2 = \frac{1}{2}(0.047)(48.0)^2 - (0.047)(9.8)(24.8) = 42.7211~J$

    Quote Originally Posted by rcmango View Post
    (b) What is its speed when it is 10.0 m below its highest point?


    not sure why b won't be given by vf^2 = vo^2 + 2*g*(yf - y0)
    (I corrected some mistakes in your equation.)

    The answer is that it does. However the v's in this formula are only the y components of the velocities in question. So you can use this equation, but once you get the y component of the final velocity, you need to add it vectorally to the x component of the final velocity, which is a difficult thing to do as we haven't been given the angle ball was hit at! (But it is quite possible to do this problem, it just takes a while.)

    Again, think of energy techniques. I'll use the same coordinate system for this part of the problem. (You can use a new one with +y upward and the 0 for GPE at the max height. Then your final y is -10 m.)
    Again we have
    $\displaystyle \Delta E = 0$

    $\displaystyle \Delta KE + \Delta PE = 0$

    $\displaystyle \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 + mgh - mgh_0 = 0$

    $\displaystyle \frac{1}{2}(0.047)v^2 - (42.7211) + (0.047)(9.8)(14.8) - (0.047)(9.8)(24.8) = 0$
    (The 42.7211 J is from part a. It is the kinetic energy of the ball when it is at max height.)

    I get (after a lot of simplification) $\displaystyle v = 44.8767~m/s$
    (This might seem to be large in comparison to the initial speed, but most of it comes from the horizontal component of the velocity, which is about 42 m/s.)

    -Dan

    Note: By the way, if you read your textbook carefully you will note that we can apply $\displaystyle v^2 = v_0^2 + 2a(s - s_0)$. The thing to note is that this is a vector equation. The form of this equation for more than 1D motion is
    $\displaystyle v^2 = v_0^2 + 2\vec{a} \cdot ( \vec{s} - \vec{s_0} )$
    where the s's are displacement vectors and the $\displaystyle \cdot $ between the vectors is the "dot product" or "scalar product" depending on how you learned it. (And I have defined $\displaystyle v^2 = \vec{v} \cdot \vec{v}$ as is typically done.)
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