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1 September, 10:28

A small first-aid kit is dropped by a rock climber who is descending steadily at 1.3 m/s. After 2.5 s, what is the velocity of the first-aid kit, and how far is the kit below the climber?

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  1. 1 September, 12:21
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    D = 30.625 m

    Explanation:

    given,

    Speed of the climber = 1.3 m/s

    time = 2.5 s

    acceleration due to gravity = 9.8 m/s²

    initial speed of the kit = 1.3 m/s

    velocity of the kit after 2.5 s

    using equation of motion

    v = u + a t

    v = 1.3 + 9.8 x 2.5

    v = 25.8 m/s

    distance travel by the kit in 2.5 s

    v² = u² + 2 g h

    25.8² = 1.3² + 2 x 9.8 x h

    19.6 h = 663.95

    h = 33.875 m

    distance travel by the rock climber in 2.5 s

    distance = speed of climber x time

    h' = 1.3 x 2.5

    h' = 3.25 m

    Distance between kit and rock climber

    D = h - h'

    D = 33.875 - 3.25

    D = 30.625 m

    The kit is 30.625 m below climber.
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