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17 December, 06:32

The distance a body falls from rest varies directly as the square of the time the body is falling. If a body falls 81 ft. in 3 seconds, how far will it fall in 7 seconds?

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  1. 17 December, 06:46
    0
    You are saying that X = k*t^2 where k is a constant

    Plugging in the 64 ft and 2 seconds k = 16 so,

    X = 16*49 = 784 ft.
  2. 17 December, 06:53
    0
    Let the distance = D

    Let the time taken = T

    From the question, we are told that distance is directly proportional to the square of the time the body takes to fall,

    that is, D = K * [T^2]

    K is a constant and we have to find its value;

    Using the value given in the question;

    D = 81 feet

    T = 3 seconds

    From our original equation,

    K = D / [T^2] = 81/3^2 = 81/9 = 9.

    Therefore, K = 9

    To find the distance through which the body fall in 7 seconds, we have,

    K = 9

    T = 7 seconds

    D = K * [T^2]

    D = 9 * [7*2] = 9 * 49 = 441.

    Therefore, the body will fall through a distance of 441 feet in 7 seconds.
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