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Projectile Motion with a Spring!

  1. May 23, 2004 #1
    Im having a little trouble deriving a single equation that will predict the travel distance of a steel ball launched by a horzontal spring launcher. The spring launcher doesnt launch at an angle. Any help on this would be appreciated. Thanx
  2. jcsd
  3. May 23, 2004 #2


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    Define the system completely.
    Is the launcher some height above the ground ?
    Do you know the spring constant ?
    Do you want to find the distance as a function of the compression of the spring ?

    Assuming answers are YES, YES and YES...the following should work approximately :

    Range = d*SQRT(2hk/mg),
    where d : size of spring compression = uncompressed length - compressed length
    h : height of launcher from ground
    k : spring constant, and
    m : mass of the steel ball

    You will probably find, that the real range will be less than the above predicted value, because
    1. of air resistance, and
    2. some residual energy in the spring after launch

    Both these corrections can be calculated for a more accurate answer.
  4. May 23, 2004 #3


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    A more accurate answer can be had by doing a simple test lanch first. Point your launcher vertically upwards and measure the maximum height reached by the ball from the end of the spring. Call this height, H. Then the range for the horizontal launch will be :

    Range = SQRT(4*H*h)

    This removes nearly all the error from 2. above, and some of the error from 1.
  5. May 26, 2004 #4
    Firstly, i imagine you know the spring constant of the spring, if not, you can work it out using simple harmonic motion equations.

    Once you know this, you can begin to work out its acceleration as it is in contact with the spring. This can be done by using the eqations:


    (where k is the spring constant and l is the distance you pull the spring back)

    By then applying [tex]F=ma[/tex]:


    you can find the acceleration relatively easily:


    Then it would just be a case of using suvat equations:

    [tex] v^2=u^2+2as[/tex]


    once you have found v, you can use projectile methods to calculate how far it will travel:

    You will know, or be able to measure the height of the apparatus, so using suvats you can work out how long it would take to hit the floor (let the previous value of v be u now):

    [tex]s=0.5a(t^2) [/tex]




    As the value of t is common in both components, you can substitute it in for the horizontal component, hence to find how far it will travel:



    ignoring air resistance and assuming the spring is smooth. (hence an ideal situation)

    I think this is right. Although it might not be. So use it if you want.

    Yes youth, tweak
    Last edited: May 26, 2004
  6. May 26, 2004 #5


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    The problem with this approach is that you assume there is a unifom acceleration during the time the ball is in contact with the spring. This is not true. The acceleartion varies harmonically. But you can still find v by solving the equation of motion of a mass loaded spring (need damping to be accurate).
  7. May 26, 2004 #6
    To eradicate this problematic approach, i will consider the spring moving harmonically:

    here goes

    [tex]a = -\omega^2 l sin\omegaT[/tex]
    [tex]a = -\omega^2 d[/tex]
    [tex]a = -\Frac{k}{m} d[/tex]
    [tex]T = \frac{2pi}{\omega}[/tex]
    [tex]d = \frac{1}{l sin \frac{\frac{-k}{m}\frac{2}{\pi}}{\sqrt{\frac{k}{m}}}} [/tex]

    my previouse equation:


    Replace l with d:

    [tex]Range = \frac{\sqrt{\frac{4kh}{mg}}}{ - l sin \frac{2\pi\sqrt{k}}{m}}}[/tex]

    Is this right?

    I don't know. Hopefully!!!!
    Last edited: May 26, 2004
  8. Aug 25, 2005 #7
    Okay, same problem, but this time it's a tennis ball and it's launched at a 30deg. angle. In addition, the mass of the ball is 5oz./0.14kg, launch height is 10.25ft. above ground where target is 50.25ft. away horizontally. Theoretically, the ball should hit the target, but ours hit at 2'11", 7'6", 16'1", and 38'2" horizontally, with an average of 20'7".

    Need to calculate the spring constant K from this, plus a visual calculation of where to "expect" it to land, if the testing hadn't actually been done.
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