1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Free fall acceleration

  1. Sep 18, 2007 #1
    1. The problem statement, all variables and given/known data

    A bolt is dropped from a bridge under construction, falling 98 m to the valley below the bridge.

    (a) In how much time does it pass through the last 30% of its fall?


    (b) What is its speed when it begins that last 30% of its fall?


    (c) What is its speed when it reaches the valley beneath the bridge?


    2. Relevant equations



    3. The attempt at a solution
    Tried splitting the problem into 2 parts by multiplying 98 x .3 to solve for the final velocity for the first 70% of the fall, then using that velocity as the initial velocity for the 30% of the fall and the other 30%(29.4m) of the trip.

    [tex]\Delta[/tex]t=-vi +/- [tex]\sqrt{vi^{2}+2ay}[/tex] / a
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 18, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    That's a good approach. What answers did you get?
     
  4. Sep 18, 2007 #3
    got 8.20 for part a seconds but it was wrong. so i stopped at part a
     
  5. Sep 19, 2007 #4

    learningphysics

    User Avatar
    Homework Helper

    Can you show what you did to get that?
     
  6. Sep 19, 2007 #5
    vf[tex]^{2}[/tex]=0 + 2(-9.8m/s^2)(-68.6m)
    vf= 36.69m/s

    [tex]\Delta[/tex]t = -36.69m/s +/-[tex]\sqrt{36.69m/s^2 + 2(-9.8m/s^2)(-29.4) }[/tex] /-9.81

    =8.20s
     
  7. Sep 19, 2007 #6

    learningphysics

    User Avatar
    Homework Helper

    Everything looks right except, your denominator in your quadratic solution... I think it should be 9.81, not -9.81.... so you'll chose the positive numerator instead of the negative one...
     
    Last edited: Sep 19, 2007
  8. Sep 19, 2007 #7
    why 9.81 when gravity is always negative?


    that means the time would actually be 0.723? i dont know if that sounds logical
     
    Last edited: Sep 19, 2007
  9. Sep 19, 2007 #8

    learningphysics

    User Avatar
    Homework Helper

    Can you write out the quadratic equation you had?
     
  10. Sep 19, 2007 #9
    [tex]\Delta[/tex]t=-vi +/- [tex]\sqrt{vi^{2}+2ay}[/tex] / a
     
  11. Sep 19, 2007 #10

    learningphysics

    User Avatar
    Homework Helper

    Ah... I think I see now.

    Here you should have taken vf as -36.69m/s, not +36.69m/s. ie the negative square root is what you want.
     
  12. Sep 19, 2007 #11

    learningphysics

    User Avatar
    Homework Helper

    Yes, that equation is correct.
     
  13. Sep 19, 2007 #12
    omg it was right! thanks!

    now to try and answer part b and c.
     
  14. Sep 19, 2007 #13

    learningphysics

    User Avatar
    Homework Helper

    cool! another way to solve that problem that I just saw... you can get the total time of the drop. and subtract the time of the first 70%

    So solving (1/2)gt^2 = 98, gives t = 4.4699s. And the solve the time for the first 70%. (1/2)gt^2 = 0.7*98, gives t = 3.7397s, so the difference is the time you need 0.730s.
     
  15. Sep 19, 2007 #14
    i got both b and c.

    b - 36.69
    c - 43.85


    thanks learningphysics for getting me started on that problem.
    Much appreciation.
     
  16. Sep 19, 2007 #15

    learningphysics

    User Avatar
    Homework Helper

    cool! no prob. good job!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Free fall acceleration
  1. Free Fall- acceleration (Replies: 11)

  2. Free fall acceleration (Replies: 2)

Loading...