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A steel ball rolls with constant velocity across a tabletop 0.950m high. IT rolls off and hits the ground +0.352m horizontally from the edge of the table. How fast was the ball rolling?

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5 answers

  1. y=vit+1/2at^2
    -9.5=1/2(-9.81)t^2
    t=0.44 s

    x=vit
    .352=vi(.44)
    vi=0.8 m/s

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  2. 8.0

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  3. Get the time to fall from the height, H = 0.950 m. Call that time T. You will need to use the acceleration of gravity (g) to compute T.
    H = (g/2) T^2
    Haven't you seen that equation before for falling bodies?

    Then use the relationship
    V T = 0.352
    to get V

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  4. 1.35 m/s

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  5. A steel ball rolls with constant velocity across a tabletop 0.950m high. IT rolls off and hits the ground +0.352m horizontally from the edge of the table. How fast was the ball rolling?

    GIVEN:

    Vi = 0 m/s
    d = 0.950 m ---> 0.352 m
    a = 9.81 m/s^2
    t = ?

    d = Vit + 1/2 at^2

    0.352 = 1/2 (9.82 m/s^2) (t)^2

    0.352 = 4.905t^2

    --> Divide by 4.095

    square root - 0.071763507 = t^2

    0.26 s = t

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