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Derive a general formula for finding range with a spring launcher

  1. Jun 11, 2009 #1
    1. The problem statement, all variables and given/known data
    So I made a spring launcher that will fire marbles. I need to derive a general formula for finding range for any given angle and x value.


    2. Relevant equations
    1) v22=v12+2ad
    2) Ek=Es
    3) 1/2mv2=1/2kx2
    4) d=v*t
    This projectile was shot from 1.1m off of the ground.

    3. The attempt at a solution
    I have no idea where to start.
     
    Last edited: Jun 11, 2009
  2. jcsd
  3. Jun 11, 2009 #2

    Cyosis

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    This is an experiment you're doing at home? The equations you've noted down do not take resistances into account. Marbles are relatively heavy though so you should be alright with these equations to some extend.

    Using equation 3) you can calculate the velocity with which the marble leaves the spring, you can then split v up in its x and y components. When it leaves the spring it's trajectory will roughly be a parabola. The x-distance is given by your equation 4. The equation that is missing is the one that describes the motion in the y-direction. Can you come up with an equation for the y-direction.

    hint: what is the only force on the object and in which direction does it point?
     
  4. Jun 11, 2009 #3
    Yes it is an experiment I am doing at home and I am not too worried about resistance.
    Ok, I am going to assume it is
    v2=v1+at because I need time, and I don't have distance.
    t=v1sinθ/a because v2 is 0 halfway up
    so
    t=2(vsinθ/a) still am not accounting for the 1.1m off the ground (not sure how to do that)

    1/2mv2=1/2kx2
    v=((kx2)/m)^1/2 but I only want the x speed so
    v=(((kx2)/m)^1/2)cosθ

    So
    d=v*t
    d= (((kx2)/m)^1/2)cosθ * 2(vsinθ/a)

    Still does not accounting for the 1.1m off the ground
     
    Last edited: Jun 11, 2009
  5. Jun 11, 2009 #4

    rock.freak667

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    For you to account for the 1.1m you will need to use the formula y=y0+uyt-1/2gt2

    then to get the time for the entire motion, set y=0 and use the quadratic equation formula for t.
     
  6. Jun 11, 2009 #5

    Cyosis

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    To clear somethings up what is this range you want to find?

    Not true since you start at 1.1 meters and the projectile lands at 0 meters. The general kinematic equation you're looking for is [itex]s=s_0+v_0t+\frac{1}{2}a t^2[/itex].

    Edit:kept the post open for way too long, not adding anything to rock's explanation.
     
  7. Jun 11, 2009 #6
    I am looking for horizontal range.
    My equation sheet has the formula
    dy=v2t-1/2at2

    what does y0/s0 stand for?
     
    Last edited: Jun 11, 2009
  8. Jun 11, 2009 #7

    rock.freak667

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    displacement.

    There should also be a t in the formula for dy.

    BUT your equation assumes you started with 0 distance (vertically since you said dy)
     
  9. Jun 11, 2009 #8
    Sorry about the t.

    if s0 stands for displacement (it should be -1.1 then?), then I set s = 0 and do the quadratic?

    EDIT: Also why is yours +1/2at^2 where as mine and cyosis is -1/2at^2 ?
     
  10. Jun 11, 2009 #9

    rock.freak667

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    well it can be -1.1 but we took the origin to be where the marble will land, 1.1m below where it was launched.

    well the general form is

    s=s0+ut+1/2at2

    Cyosis put a - sign because the acceleration is due to gravity which acts downwards, which is usually taken as -ve.
     
  11. Jun 11, 2009 #10
    Ok so I did:
    dy=v1yt+1/2ayt2
    -1.1=v1yt-4.905t2 9.81 for g here.
    4.905t2-v1yt-1.1

    t=-v1y+/-(((-v1y)2-4(4.905)(-1.1))^.5)/2(4.905)
    t=-vsinθ+(((-vsinθ)2+21528)^.5)/9.81 (has to be positive or time will be negative)

    then from before:
    1/2mv2=1/2kx2
    v=((kx2)/m)^1/2 but I only want the x speed so
    vx=(((kx2)/m)^1/2)cosθ

    now put those both in dx=vx*t

    dx=(((kx2)/m)^1/2)cosθ * (-vsinθ+(((vsinθ)2+21.582)^.5)/9.81)

    dx=(((kx2)/m)^1/2)cosθ * (-(((kx2)/m)^1/2)sinθ+(((((kx2)/m)sinθ)+21.582)^.5)/9.81)

    not sure if I can simplify further or if I made a mistake.
     
    Last edited: Jun 11, 2009
  12. Jun 11, 2009 #11

    rock.freak667

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    You should get

    [tex]t= \frac{vsin\theta+\sqrt{v^2 sin^2 \theta-4(\frac{1}{2}g)(-1.1)}}{g}[/tex]


    then put that into dx=(vcosθ)t


    (sin2θ=2sinθcosθ)
     
  13. Jun 11, 2009 #12
    Can you show all your work on how you got t, because I keep getting a negative in front.

    also fixed my final t line.
     
  14. Jun 11, 2009 #13

    rock.freak667

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    [tex]-1.1=(vsin\theta)t-\frac{1}{2}gt^2[/tex]

    [tex]\Rightarrow \frac{1}{2}gt^2-(vsin\theta)t-1.1=0[/tex]

    [tex]a= \frac{1}{2}gt^2 \ b= -vsin\theta \ c=-1.1[/tex]

    [tex]t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
     
  15. Jun 11, 2009 #14
    Thanks
    I am going to assume here g = 9.81 not -9.81 or else the inside of the root could fail.
     
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