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Need help before quiz today. Kinematics

  1. Sep 18, 2008 #1
    1. The problem statement, all variables and given/known data
    An elevator accelerates from the ground with a uniform acceleration 'a.' After 4.0 seconda, an object is dropped out an opening in the floor of the elevator and that object hits the ground 4.5 seconds later.

    a) what is the value of the acceleration 'a?'

    b) How high was the elevator when the object was dropped?

    c) What is the speed of the object when it hits the ground?


    2. Relevant equations
    V=Vo+at
    X=Xo+Vot+1/2(a)(t^2 )
    V^2=Vo^2+2a(x-Xo)



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 18, 2008 #2

    LowlyPion

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    As it drops out the elevator what speed is the object going up? Because the object will continue upward until returning by gravity.

    Develop equations from the variables you know to find those you don't. That's always the way.
     
  4. Sep 18, 2008 #3
    V(elevator) = 0+a(4.0)???
    i don't understand how to do it, thats why i posted on here lol could you show the solution for this one so i understand how it works? :/
     
  5. Sep 18, 2008 #4

    LowlyPion

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    Ok that's a start.

    Now how much longer until it hits? That's given as 4.5 seconds.

    Now what is the height it was released from? H = + 1/2 a t2 = 1/2 a (4)2 = 16/2 * a = 8a

    It's getting easier and easier to get to the solution.

    So to model the x position of the object we know it is originally 8*a, to which we add +Vo*(t)
    and the subtract the effect of gravity - 1/2*g*t2

    X = 0 = 8*a +(4a) * (4.5) - 1/2*g*(4.5)2

    We set X = 0, it hits the ground 0, because we want to solve for "a".

    Looks like that yields the "a" that you need to solve for everything.
     
    Last edited: Sep 18, 2008
  6. Sep 19, 2008 #5
    please correct me if I'm incorrect, this is just my attempt..

    dist. travelled by the elevator the moment the ball falls = dist. travelled by the ball when it hits the ground

    Since the ball takes 4.5 seconds to reach the ground, under the effect of gravity, the ball would have travelled 1/2 * a * t² = 1/2 * 9.8 * 4.5² = 99.225 m

    So, 99.225 = 1/2 * a * 4²
    a = 12.4 ms-2

    (a) 12.4ms-2

    (b) 99.225m

    (c) v = at = 44.1 ms-1
     
    Last edited: Sep 19, 2008
  7. Sep 19, 2008 #6

    LowlyPion

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    Sorry but that answer is incorrect because you neglected to account for the upward velocity of the ball when released.
     
  8. Sep 19, 2008 #7
    thanks for pointing out my mistake, so does it mean that if an object is travelling upward with certain velocity, and is suddenly released, it doesn't obey the normal free-fall rule? (assume air resistance is neglected)
     
  9. Sep 19, 2008 #8

    LowlyPion

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    It acts of course, at that point, as any object in a uniformly accelerated field. When you know the initial velocity up and the height it is released, then you can relate that to the time to drop.

    My earlier post suggests a path to solution.
     
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