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Projectiles are sweet

  1. Mar 24, 2007 #1
    1. The problem statement, all variables and given/known data
    A battle ship (4.09*107 kg) fires a salvo of 3 rounds from its foward turret in the direction of the bow at an angle of 10 degrees above the horizontal. Each projectile weighs 1220 kg, each barrel is 20.9 m long and the muzzle velocity (I don't know whether this is initial or final velocity) of the projectile is 770 m/sec.

    What is the acceleration of each projectile in the barrel and how long does it take to travel down the barrel?

    2. Relevant equations
    Vf=Vi + a(T)
    S= Vi(T) + 1/2(a)(T)^2
    a= vf-vi/T

    3. The attempt at a solution

    no clue!!
    Last edited: Mar 24, 2007
  2. jcsd
  3. Mar 24, 2007 #2
    I would assume that the muzzle speed is that of the projectile at the end of the barrel.
    What speed does the projectile start with?
  4. Mar 24, 2007 #3
    projectile starts with a speed of 770 m/sec.
  5. Mar 24, 2007 #4
    You want to determine the acceleration of the projectile while in the muzzle, right? It seems the only reasonable thing to assume is that the projectile attains the "muzzle speed" at the end of the muzzle, at least to me.

    *edit* and yes they are sweet.:biggrin:
    Last edited: Mar 24, 2007
  6. Mar 24, 2007 #5
    Here's what I've concocted so far. However, I'm not sure if it's correct, for the acceleration comes out to be negative and i don' think i could get a negative acceleration for this type of problem:

    a= -770/t

    a= -770/.0543
    a= -14180 m/sec^2

    S=Vi(T) + 1/2(a)(T)^2
    20.9=770(T) + 1/2 (-770/T)(T)^2
    20.9=770(T) - 385(T)
    T=.0543 sec
  7. Mar 24, 2007 #6
    To me it looks like you are stating that the initial speed is 770 and the final speed is 0. That doesn't make sense to me.
  8. Mar 24, 2007 #7
    so would it be the other way around?? initial velocity is 0 and final velocity is 770??


    a=770/.018= 42777 m/sec^2

    S= Vo(T) + 1/2(A)(T)^2
    20.9=770(T) + 1/2 (770/T)(T)^2
    20.9= 1155(T)
    T= .018 sec

  9. Mar 24, 2007 #8
    This is right, but perhaps a more straightforward approach assumes vf=770, and Vi=0,

    Given that Vf^2-Vi^2=2ax, then a=770^2/(2*20.9)
    a=14184m/s^2 and t would just be the distance divided by ave velocity:
    20.9/385=0.0543s Different ways to skin a cat. Best to have many knives.
  10. Mar 24, 2007 #9
    Yes the solution is the same mathematically, but physically, at this level, hmmmm?
    If the OP recognizes that both problems are the "same," then great!
  11. Mar 24, 2007 #10
    me and denverdoc contrived totally different answers. So which one would be right? I'm not sure if my calculations from my previous post were valid anyway.
  12. Mar 24, 2007 #11
    Oh, sorry didnt catch your last post. You now should consider the initial speed of the projectile in your distance equation.
  13. Mar 24, 2007 #12
    HA! I got it! thank you both, robb_ and denverdoc.
  14. Mar 24, 2007 #13
    : )
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