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5 December, 23:56

A child, starting from rest at the top of a playground slide, reaches a speed of 7.0 meters per second at the bottom of the slide. what is the vertical height of the slide? [neglect friction.]

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  1. 6 December, 00:05
    0
    We know that the child will follow a accelerated movement due the gravity, according these equations:

    h = v0 * t + 1/2 * g * t^2

    v = v0 + g*t

    where h is height, v is final speed, v0 is initial speed, g is gravity constant equal to 9.8 m/s2 and t is time

    So the time to reach the bottom of the slide

    v = v0 + g*t = > 7 = 0 + 9.8*t

    t = 7/9.8 = 0.714 seconds

    And then the height is:

    h = v0*t + 1/2*g*t^2

    h = 0*0.714 + 1/2*9.8*0.714^2 = 2.5 meters

    So the height of the slide is 2.5 meters
  2. 6 December, 00:16
    0
    for child sliding down;

    at h the child’s potential energy: PE = m*g*h

    at the bottom of the slide the kinetic energy: KE = (1/2) * m*V^2

    m * g * h = (1/2) * m * v^2

    g * h = (1/2) * v^2

    h = (1/2) * (v^2) / g

    h = (1/2) * (7^2) / 9.8

    h = 2.5 meters
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