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!

The Cliff Diver - Projectile motion

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data


    A swimmer dives off a cliff with a running horizontal leap. What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom which is 1.5 m wide and 9.5m below the top of the cliff.



    2. Relevant equations

    X=V0xT Y=-1/2gt^2

    3. The attempt at a solution

    I have X=1.5 and Y=9.5

    9.5=-1/2*9.8*t^2
    1.5=V0x*t

    From there I am lost, Ive tried squaring T and so on. Any help is appreciated.
     
  2. jcsd
  3. Feb 2, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    Solve the first equation for the time. Substitute it into the second to find the Vo.
     
  4. Feb 2, 2009 #3
    9.5=-1/2*9.8*t^2
    9.5=-4.9*t^2
    1.93=t^2
    t=1.389

    1.5=Vox*1.389
    1.08=Vox

    Does this sound right?
     
  5. Feb 2, 2009 #4

    Delphi51

    User Avatar
    Homework Helper

    Looks good!
    In case your marker is picky, it would be a good idea to deal with the extra minus sign in the first part. Either think of "down" as positive and not use any minus signs, or put a minus sign on the 9.5 to indicate it is down.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: The Cliff Diver - Projectile motion
  1. Diver Proj. Motion (Replies: 3)

Loading...