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Math Help - Dynamics question involving impulse momentum

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    Dynamics question involving impulse momentum

    The question is:

    A 5oz baseball is 3ft above the ground when it is struck by a bat. The horizontal distance to the point where the ball strikes the ground is 180ft. Photographic studies indicate that the ball was moving approximately 100ft/s before it was struck, the duration of the impact was 0.015s, and the ball was traveling at 30 degrees above the horizontal after it was struck. What was the magnitude of the average impulsive force exerted on the ball by the bat?

    So I used kinematics to find the velocity of the baseball right after impact, utilizing the distance it traveled. I found it to be 112.48ft/s, whether that's correct or not I have no idea. So using that, I used the formula (t2-t1)SUM(Fbat)=MV2-MV1. I found the answer to be 2029.3i + 1172j - 6666.7. I'm thinking I did something wrong. And wow, someone seriously just cropdusted my area of the library
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  2. #2
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    Quote Originally Posted by pakman View Post
    The question is:

    A 5oz baseball is 3ft above the ground when it is struck by a bat. The horizontal distance to the point where the ball strikes the ground is 180ft. Photographic studies indicate that the ball was moving approximately 100ft/s before it was struck, the duration of the impact was 0.015s, and the ball was traveling at 30 degrees above the horizontal after it was struck. What was the magnitude of the average impulsive force exerted on the ball by the bat?

    So I used kinematics to find the velocity of the baseball right after impact, utilizing the distance it traveled. I found it to be 112.48ft/s, whether that's correct or not I have no idea. So using that, I used the formula (t2-t1)SUM(Fbat)=MV2-MV1. I found the answer to be 2029.3i + 1172j - 6666.7. I'm thinking I did something wrong. And wow, someone seriously just cropdusted my area of the library
    Impulse: \Delta \vec{p} = m(\vec{v}_\text{final} - \vec{v}_\text{initial}{\color{red})}.

    You know \vec{v}_\text{initial}.

    Let U be the speed of the ball after being struck by the bat. Then the vertical and horizontal components of \vec{v}_\text{final} are U \sin (30^0) and U \cos (30^0) respectively.

    Making the usual assumptions (negligible air reistance, no crazy spin on ball etc.) you can apply the straight line motion formula for constant acceleration in the vertical and horizontal directions of the balls motion:

    Horizontal direction: x = 180 ft, u = U \cos 30^0 = ...... ft/sec, a = 0, t = ?

    Substitute the data into x = ut + \frac{1}{2} at^2 to get equation (1).

    Vertical direction: x = 0, u = U \cos 30^0 = ...... ft/sec, a = -32 ft/sec^2, t = ?

    Substitute the data into x = ut + \frac{1}{2} at^2 to get equation (2).

    Solve equations (1) and (2) simultaneously for U. Use this value to get the components of and hence \vec{v}_\text{final}. I'm too lazy to check if the value of U agrees with your value .....

    So now you can write down the impulse.
    Last edited by mr fantastic; April 20th 2008 at 11:32 PM. Reason: Added the red bracket.
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