Just as the last passenger steps out of the elevator at the thirteenth floor, a malfunction occurs and it begins dropping toward the ground

Question

Just as the last passenger steps out of the elevator at the thirteenth floor, a malfunction occurs and it begins dropping toward the ground floor in free fall, Ignoring friction (and the fact that elevators are designed to avoid this type of accident), what is its downward acceleration? How fast is it moving after 2.5 s?

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Xavia 5 years 2021-09-04T02:43:06+00:00 1 Answers 74 views 0

Answers ( )

  1. binhan
    0
    2021-09-04T02:44:29+00:00
    Reply

    +10m/s²–25m/s

    Explanation:

    it’s downward acceleration is +10ms-²

    since it is falling freely under the influence of gravity

    Since the elevator drops downward, that is initially from rest , u=0

    so for this equation S=u + ½at²

    but since u is 0 the new equation becomes S=½at²

    thus from the new equation we determine the height attained after 2.5s during the fall

    substituting 2.5s in the equation…

    we get s to be 31.25m

    then we substitute s in this equation when u=0 to get the velocity of the body

    v²=u²+2as

    v²=2as

    v²=2(+10•31.25)

    v²=625

    v=25m/s

    i suppose this will help

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