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Spring launcher equation for x so it hits the target

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data
    I need to find out how far to pull the spring back so it will travel to the target. The launcher and the target is both set on the floor. The spring is the projectile. I will be given the distance on the day of testing. i should take into account air resistance so i will be more accurate.

    Mass of spring = 0.024kg
    k = 25N/m
    launch angle set at 40 degrees
    length of spring when not stretched is 0.20m
    height of launcher is 0.54m
    My trials pulling spring back and recording distance
    .25m went .52m
    .3m went 1.67m
    .34m went 2.2m
    .4m went 4.78m
    .43m went 5.37m

    2. Relevant equations
    conservation of energy
    Es = Ek
    1/2mv^2 = 1/2kx^2
    projectile motion
    d = v1xt
    v1x = v1cosӨ

    3. The attempt at a solution
    1/2mv^2 = 1/2kx^2
    i get stuck on this part. plugging in numbers but i do not have the velocity. What i do next?
     
  2. jcsd
  3. Jan 30, 2013 #2

    haruspex

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    Is the target at floor level?
    You need an equation relating to vertical motion, relating initial speed, acceleration, time and height difference.
     
  4. Jan 30, 2013 #3
    yes it's on the floor
     
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