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Projectile motion ball question

  1. Apr 10, 2009 #1
    A ball moving with speed v rolls off a shelf of height h and strikes the floor below a distance x1 from the edge. A second ball moving with the same speed rolls of a self of height 2h and strikes the floor a distance x2 away from the edge. Determine the ratio of distances x1/x2.



    [tex]\Delta[/tex]x=Vxt
    [tex]\Delta[/tex]y=vt +1/2st2



    The horizontal motion of the ball is constant and, therefore, not affected by the height from which it rolls. But I cannot figure out how to formulate the ratio of the distances. I know that time relates the two equations of motion and that because the ball is rolling it has no initial vertical velocity and that since it is free fall the acceleration would be the same for the two equations. I just can't seem to get the equation to make sense when I write it out. Any help would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Apr 10, 2009 #2

    LowlyPion

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    Well I can see you are almost grasping it.

    The horizontal motion is constant, that's true, but what affects the distance at which it lands?

    d1 = V*t1
    d2 = V*t2

    So ... the ratio of the distances then will be in the same ratio as the ratio of the times won't it?

    And now is there some way to relate the ratio of the times using the heights?
     
  4. Apr 12, 2009 #3
    I think I have it now.

    2y = 1/2 at12 divided by y = 1/2 at22

    gives me 2= t12t22

    so t1t2 =square root of 1/2.

    Thanks for your help, still not entirely sure I understand what I did. I guess that's what practice is for.
     
  5. Apr 12, 2009 #4

    rl.bhat

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    Check this step.
     
  6. Apr 12, 2009 #5

    LowlyPion

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    I think you really meant
    2= t12/t22

    Which yields the ratio

    t1/t2 = √2

    But I would note the original statement has y2 being 2*y1, so I think you have your subscripts reversed.

    The higher ball would fall farther from the base by a factor of √2 is the sense of you should be grasping.
     
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