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!

Kinetic's Problem

  1. Apr 10, 2005 #1
    If a sprinter running at a speed of 10 meters per second could convert his/her kinetic energy into upward motion, how high could he/she jump?

    I understand that KE= 1/2 mass X (Speed)2 But im really not sure how I could do this without a mass.
     
  2. jcsd
  3. Apr 10, 2005 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    At the point where all is kinetic energy has been converted in potential energy.. how high will he be?
     
  4. Apr 10, 2005 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    U don't need the mass of the sprinter in this elementary setup.

    Daniel.
     
  5. Apr 10, 2005 #4
    Well this elementary setup doesn't seem so elementary to me. The point where his potential energy turns into kinetic energy will be the first instance he is in motion.
     
  6. Apr 10, 2005 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    It's the other way around.He's high-jumping.Which means that his KE converts into PE and not viceversa.

    Daniel.
     
  7. Apr 10, 2005 #6
    I feel stupid now. so if he used the same amount of energy he would be able to jump 10 meters high?
     
  8. Apr 10, 2005 #7

    Doc Al

    User Avatar

    Staff: Mentor

    No. Figure it out. Set the initial KE equal to the final gravitational PE. Then you can solve for the maximum height.
     
  9. Apr 10, 2005 #8
    This is where I am lost the only formulas that the book provides is PE=mgh and KE= 1/2mv*

    So i dont see how i can solve this PE=mgh=1/2m(100)=KE
     
  10. Apr 10, 2005 #9

    Doc Al

    User Avatar

    Staff: Mentor

    That's the correct equation: [itex]mgh = 1/2 m v^2[/itex]; now just solve for h. (Divide both sides by mg!)
     
  11. Apr 10, 2005 #10
    I think if I would of realized that mass and weight are directly proportional sooner I would of had less trouble with this. So the hight would equal 50 meters?
     
  12. Apr 10, 2005 #11

    Doc Al

    User Avatar

    Staff: Mentor

    [itex]h = v^2/(2g)[/itex]. Plug in the numbers: v = 10 m/s; g = 9.8 m/s^2.
     
  13. Apr 10, 2005 #12
    Ok so one meter. Now the question is how did you get this formula and where did you get g=9.8 from. Wouldn't mgh=1/2mv* come out to h=1/2v*? I understand that 9.8 is that rate that things fall but im not sure how this plugs into all this. Isn't mg that weight of the object?
     
  14. Apr 10, 2005 #13

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    U mean ~5 m...No,it wouldn't come out to what u've written.You can simplify only through "m"...

    Daniel.
     
  15. Apr 10, 2005 #14
    I didn't see the 2g in there so it's .5 meters. So when simplying you would dived 1/2m by mg to get 2g right?
     
  16. Apr 10, 2005 #15

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    It's 100 divided by 19.2.

    Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Kinetic's Problem
Loading...