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1 October, 20:29

A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t=18 s. What is the value of v?

Answers (1)
  1. J
    2 October, 00:18
    0
    73.5 m/s

    Explanation:

    The position of the first ball is:

    y = y₀ + v₀ t + ½ at²

    y = h + (0) (18) + ½ (-9.8) (18) ²

    y = h - 1587.6

    The position of the second ball is:

    y = y₀ + v₀ t + ½ at²

    y = h + (-v) (18-6) + ½ (-9.8) (18-6) ²

    y = h - 12v - 705.6

    Setting the positions equal:

    h - 1587.6 = h - 12v - 705.6

    -1587.6 = - 12v - 705.6

    1587.6 = 12v + 705.6

    882 = 12v

    v = 73.5

    The second ball is thrown downwards with a speed of 73.5 m/s
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