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13 March, 17:15

The mass of the Earth is 5.98 * 1024 kg. A 11 kg bowling ball initially at rest is dropped from a height of 2.63 m. The acceleration of gravity is 9.8 m/s 2. What is the speed of the Earth coming up to meet the ball just before the ball hits the ground? Answer in units of m/s.

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  1. 13 March, 17:32
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    v = 7.18_m/s

    Explanation:

    Velocity of the earth towards the ball is = velocity of the ball moving towards earth

    For object in free fall, we have

    Where

    v = final velocity

    u = initial velocity

    g = acceleration due to gravity

    t = time

    S = height of ball above ground

    v^2 = u^2 - 2*g * (-S)

    = 0 + 2*9.8*2.63 = 51.55_m^2/s^2

    Velocity of the ball just before it hits the ground is

    v = 7.18_m/s
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