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!

Find the time of its fall

  1. Feb 2, 2007 #1
    I have been having some problems with this question:

    An object falls from a height h from rest. If it travels a fraction of the total height of 0.5665 in the last 1.00 s, find the time of its fall.

    I started by saying h=(0.5)(g)(T^2)
    Then I said that 1-.5665h=(0.5)(g)(T-1)^2

    I have no idea if this is even remotly right. Please help.
     
  2. jcsd
  3. Feb 2, 2007 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You have the correct equation. All you need to do is set T=1 and you know that the distance travelled is 0.5665h. Then you can solve for the height put it back in the original equation and solve for total time.
     
  4. Feb 2, 2007 #3
    so that it states

    h=(.5)(9.81)(1)?????
     
  5. Feb 2, 2007 #4

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    you would have [tex] 0.5665 h =\frac{1}{2}g[/tex]

    Now rearrange for h.

    When you find h put it back into the original equation and solve for t.
     
  6. Feb 2, 2007 #5
    i got an h=8.66
    Then i put that in and got a value
    t=1.33s

    That is not the right answer though.....where am i going wrong???
     
  7. Feb 2, 2007 #6

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No worries its my fault. Give me a minute (doing many things at once) and I'll correct my mistake.
     
  8. Feb 2, 2007 #7
    know the feeling....not a problem
     
  9. Feb 2, 2007 #8

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Ok so the equation that describes the motion is:

    [tex] s=h-\frac{1}{2}gt^2 [/tex]

    Now the question tells you that in the final second the particle falls [itex]s=0.5665h [/itex] therefore,

    [tex] 0.5665h = h-v_t-\frac{1}{2}g [/tex]

    where vt is the velocity at t-1 seconds. You can work out this velocity and should get a quadratic in h.
     
  10. Feb 2, 2007 #9
    so how would I find the vt though?
    would i solve the first equation for h and then plug that into the second?
     
  11. Feb 2, 2007 #10

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What about other kinematic equations? Do you know any more?
     
  12. Feb 2, 2007 #11
    my prof does not teach? have not be given any others yet
     
  13. Feb 2, 2007 #12

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  14. Feb 2, 2007 #13
    Thank You For All The Help!!!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?