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10 April, 22:58

At a particular instant, a hot air balloon is 270 m in the air and descending at a constant speed of 3.5 m/s. At this exact instant, a girl throws a ball horizontally, relative to herself, with an initial speed of 25 m/s. When she lands, where will she find the ball? Ignore air resistance. (Find the distance, in meters, from the girl to the ball.)

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  1. 11 April, 00:33
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    Refer to the diagram shown below.

    Neglect wind resistance.

    g = 9.8 m/s².

    Quantities are measured as positive upward.

    When the ball is launched, it has a vertical velocity of - 3.5 m/s and a horizontal velocity of 25 m/s.

    The time, t, for the ball to reach the ground is

    (-3.5 m/s) * (t s) - 0.5 * (9.8 m/s²) * (t s) ² = - (270 m)

    -3.5t - 4.9t² = - 270

    4.9t² + 3.5t - 270 = 0

    t² + 0.71433t - 55.102 = 0

    Solve with the quadratic formula.

    t = 0.5[-0.71433 + / - √220.918] = 7.0745 s or - 7.788 s.

    Reject negative time.

    The horizontal distance traveled is

    d = (25 m/s) * (7.0745 s) = 176.8625 m

    Answer:

    The horizontal distance from the girl to the ball will be 176.9 m (nearest tenth)
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