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Please help me. topic: ( motion with constant acceleration : free fall )

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data

    A scaffolding is rising at 1.0 m/s when at a height of 50.0 m , then the man on the scaffolding drop a can. determine:

    a. the time needed for the can to reach the ground .

    b. what is the final velocity ?


    2. Relevant equations

    note: g=-9.8 m/s
    the letter o in the equation stands for initial example : Vo( initial velocity) .

    V=Vo +(-g)t
    y=yo + Vot + 1/2(-g)T^2
    V^2=Vo^2 + 2(-g)(y-yo)
    y-yo=(v+vo)/2 (t)


    3. The attempt at a solution


    1. The problem statement, all variables and given/known data

    t = time
    y=position
    v = velocity
    g=gravity

    2. Relevant equations



    3. The attempt at a solution

    The illustration was illustrated by our professor .

    the attempt that i was trying to solve is i tried to get each of all the final velocity.

    guys can you help me on how to solve this ! thank you !
     

    Attached Files:

  2. jcsd
  3. Jul 9, 2011 #2

    Doc Al

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    Staff: Mentor

    That's confusing. I assume you mean that a = -g, where g = 9.8 m/s^2.

    Use this formula to solve for the time.
     
  4. Jul 9, 2011 #3
    sir yes ! a= g . and the g is equals to -9.8 m/s^2

    sir ! how many times should i compute for the time ? is it 1 time or 4 times ? base on the illustration ???
     
  5. Jul 9, 2011 #4

    Doc Al

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    If you insist on g being negative, then your formula is incorrect. (It has a -g in it.)
    Why not just once? Solve for the final time when it hits the ground. What values will you put for y, y0, and v0?
     
  6. Jul 9, 2011 #5
    sir ! our professor says that it is always negative . so he is wrong ? and the right is -g=9.8 m/s^2 ? right ? thank you ! so sir we will disregards the illustration ?? and compute the final velocity just once ?
     
  7. Jul 9, 2011 #6

    Doc Al

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    Taking up as positive, the acceleration is -9.8 m/s^2.

    (The constant g is usually taken as positive, so the acceleration would be -g.)
     
  8. Jul 9, 2011 #7
    ah i see sir ! thank you. sir so i will disregard the illustration that our prof was made ? and i will focus in solving using the given value's in the problem ?
     
  9. Jul 9, 2011 #8

    Doc Al

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    I suppose the illustration is there to help you visualize what's going on. (I think the velocity at point 3 should be negative.) But you can solve for the final time in one step, you don't need to solve for all the intermediate points.
     
  10. Jul 9, 2011 #9
    so sir. i will be focus on the formula that u was given to me ?
     
  11. Jul 9, 2011 #10

    Doc Al

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    Yes.
     
  12. Jul 9, 2011 #11

    t^2=-2(y)/g
    t^2=2(50)/9.8

    t=3.19 s


    y-50=v^2-(1.0)^2/2(9.8)
    v^2-(1.0)^2 = (-50)*2(9.8)
    v^2-1 =-980
    V^2/-1=-980/-1
    v^2=980

    V=31.30 m/s

    sir am i right ? regards on solving the problem ? thank you !
     
  13. Jul 9, 2011 #12

    Doc Al

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    :confused: Why didn't you use the formula I indicated in post #2?
     
  14. Jul 9, 2011 #13
    sir i derived the formula that you indicate like what i saw in my lecture .
     
  15. Jul 9, 2011 #14

    Doc Al

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    You posted some formulas. I quoted one and said "use this one". So use it. (The one you derived assumes that the initial speed is zero.)
     
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