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11 September, 13:45

A ball of mass 0.152 kg is dropped from a height 2.27 m above the ground. The acceleration of gravity is 9.8 m/s 2. Neglecting air resistance, determine the speed of the ball when it is at a height 0.903 m above the ground. Answer in units of m/s.

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Answers (2)
  1. 11 September, 14:06
    0
    Height to achieve a speed of 2 m/s = 0.20m

    Height to achieve a speed of 3 m/s = 0.46m

    Height to achieve a speed of 4 m/s = 0.82m

    Height to achieve a speed of 5 m/s = 1.28m

    Height to achieve a speed of 6 m/s = 1.84m
  2. 11 September, 14:13
    0
    8.44 m/s.

    Explanation:

    Change in Potential Energy = Mass x Acceleration From Gravity x H2 - H1

    Kinetic Energy = 1/2 (mass) x [ (v2) ^2 - v1^2]

    g * h = 1/2 * v^2

    (9.8) x (2.27) = 1/2 * (v) ^2

    v^2 = 2[ (9.8) x (2.27) ]

    v = 6.67 m/s

    g * delta h = 1/2 * delta v^2

    (9.8) x (2.27 - 0.903) = 1/2 * [ (v2) ^2 - (6.67) ^2]

    v2^2 = 71.2821

    v2 = 8.44 m/s
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