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20 December, 18:48

A marble rolls off the edge of a table top with a speed of 2.00 m/s. a.) What is the magnitude of its velocity 0.100 s later? b.) How far from the table does it land? The height of the table is 1.00m.

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  1. 20 December, 19:48
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    (a) 2.23 m/s

    (b) 0.9 m

    Explanation:

    h = 1 m, t = 0.1 second

    horizontal component of initial velocity, ux = 2 m/s

    vertical component of initial velocity, uy = 0

    (a) Let v be the velocity after 0.1 seconds. Its vertical component is vy and horizontal component is vx.

    The horizontal component of velocity remains constant as in this direction, acceleration is zero.

    vx = ux = 2 m/s

    Use first equation of motion in Y axis direction.

    vy = uy + g t

    vy = 0 + 9.8 x 0.1 = 0.98 m/s

    Resultant velocity after 0.1 second

    v^2 = vx^2 + vy^2

    v^2 = 2^2 + 0.98^2

    v = 2.23 m/s

    (b) Let it takes time t to land.

    Use second equation of motion along Y axis

    h = uy t + 1/2 g t^2

    1 = 0 + 1/2 x 9.8 x t^2

    t = 0.45 second

    Let it lan at a distance x.

    so, x = ux x t

    x = 2 x 0.45 = 0.9 m
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