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14 November, 05:03

A little boy is standing at the edge of a cliff 1000 m high. He throws a ball straight downward at an initial speed of 20 m/s, and it falls straight down to the ground below.

A) At a time of 6 seconds after it was thrown, how far above the ground is it?

The acceleration due to gravity is 10 m/s^2

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Answers (2)
  1. 14 November, 05:05
    0
    700 m.

    Explanation:

    Using equations of motion,

    Given:

    t = 6s

    g = 10 m/s²y

    S = vi*t + (g*t²) / 2

    = (20*6) + (10*6²) / 2

    = 300 m

    Height above the ground = Height of the cliff - height of the motion

    = 1000 - 300

    = 700 m
  2. 14 November, 05:32
    0
    700 m

    Explanation:

    Using the equations of motion,

    g = 10 m/s², H = 1000 m,

    Initial velocity, u = 20 m/s,

    final velocity, v = ?

    Total Time of fall, t = ?

    a) But at t = 6s, g = 10 m/s², y = The height the ball has fallen through.

    y = ut + gt²/2

    y = (20*6) + 10 (6²) / 2

    y = 300 m

    Height of the ball above the ground at t = 6s is (Height of the cliff - height the ball has fallen through) = 1000 - 300 = 700 m
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