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Kinematics Free fall problems part 2 help

  1. May 10, 2010 #1
    1. The problem statement, all variables and given/known data
    While riding on an elevator descending with a constant speed of 3.3 m/s, you accidentally drop a book from under your arm.

    (a) How long does it take for the book to reach the elevator floor, 1.0 m below your arm?

    (b) What is the book's speed relative to you when it hits the elevator floor?

    2. Relevant equations
    v = v0 + a*t
    x = x0 + v0 *t + 1/2*a*t^2
    x= x0 + v*t-1/2*a*t^2
    V^2 = v0^2 +2*a*(x-x0)


    3. The attempt at a solution

    I tried to solve for time by first solving for final velocity using V^2 = v0^2+2*a*(x-x0) =
    V^2 = (3.3m/s)^2 +2(-9.80m/s^2)(0-1m)
    V^2 = 30.49 m^2/S^2
    square root of 30.49 = 5.522 m/s

    v = v0 +a*t
    5.522 = 3.3 + (9.80)*t
    t = 0.2267

    part b) substract the book's velocity 5.522 - my velocity 3.3 = 2.222

    I got both answers wrong. I was wondering if someone could help me explain what i did wrong? thank you so much!
     
  2. jcsd
  3. May 10, 2010 #2

    Char. Limit

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    Gold Member

    Shouldn't your initial velocity be negative, considering that you're descending?
     
  4. May 10, 2010 #3
    Since you're dropping the book with an initial velocity, and the rest of the falling path is accelerating, the velocity at any moment can be calculated, as you know, with:
    v = v0 + at
    This, however, is the velocity. If we want the distance, we have to multiply by t (vt=s)
    v = v0 + at /*t
    s = v0t + at^2/2
    Now you can substitute s with the height of 1:
    1 = 3.33t + 9.81t^2/2
    Go on from here. If you have trouble extracting t which you need, it can be substituted:
    v = 3.33 + 9.81t
    v - 9.81t = 3.33
    t = v-3.33/9.81
    Insert it into the previous expression:
    1 = 3.33(v-3.33/9.81) + 9.81*(v-3.33/9.81)^2/2
    Hope you can work it out from here (you'll get the final v and from that the t).
    Good luck!
     
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