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A challenger at highschool physics

  1. Nov 5, 2007 #1
    Launching spring at targets (physics)?
    I have a question regarding physics.
    Its about launching a spring at various targets (bucket) at different distances and heights.

    What is the equation to find the amount of stretch from equilibrium (in the string) i.e. x. We are given the vertical height, horizontal distance, and the constant of the spring (k=40n/m)
    Any help would be appreciated. Thank you in advance.

    I don't have trouble doing it when its on a level surface, but when i have to hit a target at various heights like above 1 meter, or below 1 meter, I struggle to find the equation for the correct stretch from equilibrium.
     
  2. jcsd
  3. Nov 5, 2007 #2
    any help will be appreciated

    i know you have to work with the equation y=vt+1/2at^2 but I'm not sure to what degree if any.
     
  4. Nov 5, 2007 #3
    The potential energy in a uniform spring is U=1/2kx^2, where k is the spring constant and x is the displacement on the spring when it is compressed. This is equal to the kinetic energy (1/2*mass*velocity^2). From there it's just a matter of projectile motion equations.
     
  5. Nov 5, 2007 #4
    Thanks but

    Thats exactly it, thats the part that i seem to be having trouble with. Finding that simple projectile equation. I need to know the equation to sub in all the information such as height and angle of inclination in order to get the change in x
     
  6. Nov 5, 2007 #5
    Split it into two components.

    You should come up with equations for the horizontal motion and also for the vertical motion.

    Use trigonometry to resolve your initial velocity vector and then use your equations.

    Remember that there is no horizontal acceleration.

    The vertical acceleration is g.
     
  7. Nov 5, 2007 #6
    Just one question that plagues me is what you do when you initially launch from a bench and the target is on the floor. I have an angle of 45 degrees as my launch angle. Another confusion is which velocity is being used , initial or final. If initial would it look like vcos45 and vsin45. And when your shooting from a bench to floor do you ( distance of 1.4 meters) (and the length of the bench in height is .95m ) make it negative (i.e -.95m) Thanks for the help TYCO05
     
  8. Nov 5, 2007 #7
    OK. So it sounds like you can work it out if the take off and landing are on the same level.

    So then after this you can figure out how long it takes for the projectile to fall that extra vertical distance.

    You can use this time in your horizontal equation to find the extra horizontal distance.
     
  9. Nov 5, 2007 #8
    Is there a way you can show me this ....im not sure how you're suppose to start...I know that take off is definitely at .95 m and landing is actually .95 meters down to the floor. Mass of the spring if it helps is actually 50g. How do you figure out time is my major dilemma here. Thanks
     
  10. Nov 5, 2007 #9
    One other question, does it help if im using the conservation of energy thereom of Ee=Ek+Eg
    .5kx^2=.5mv^2+mgh. Tell me if im heading in the right direction
     
  11. Nov 5, 2007 #10
    Tycoo4, i have the total horizontal distance, its 1.4 meters from the launch site which is .95 meters on a bench. I have the launch angle and it's 45 degrees, also i have the constant of the spring at 40n/m. The mass of the spring is 50 g. Thats why im thinking about the conservation of energy thereom.

    So would it look like this

    0.93=vsin45t+.5(-9.8)t^2

    and if so how do i obtain v and t

    the x side should look like x=vcos45t but i still dont know how to get v and t

    and when i have them do i just put it in this equation
    .5kx^2=.5mv^2+mgh.
     
  12. Nov 6, 2007 #11
    I assume you meant 0.95

    but it should be -0.95. The projectile is landing below the origin.

    The time is the same as the horizontal time. That is, how long does it take for the projectile to travel 1.4m ?

    Or you could solve for t using the quadratic formula.
     
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