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Calculate acceleration of a ship fired from a cannon

  1. Jan 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Jules Verne
    In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of a cannon, called the Columbiad, with a speed of 12,000 yards/s. The Columbiad is 900 ft long, but part of it is packed with powder, so the spaceship accelerates over a distance of only 700 ft. Estimate the acceleration experienced by the occupants of the spaceship during launch.

    v0 = 10,975.60976 m/s
    a = ?
    d = 213.4146341 m

    2. Relevant equations

    1) [itex]x= x_0+v_0t+\frac{1}{2}at^2[/itex]
    2) [itex]v^2 = v_o^2 + 2a\Delta x[/itex]

    3. The attempt at a solution
    I tried solving the second equation, but I got an insane number that isn't plausible. My main problem is what should I consider V0, is it 0 or is it the 10,975 m/s? If I set it to 0, then i'm only calculating the acceleration from at rest to as soon as it leaves the cannon's barrel. Otherwise, if I set it to the 10,975 m/s, then I don't know the final velocity after accelerating for 213.4 m. Also, would the acceleration be considered as [itex]-9.8\frac{m}{s^2}[/itex]?
    Last edited: Jan 23, 2015
  2. jcsd
  3. Jan 23, 2015 #2


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    Staff: Mentor

    The initial velocity is zero, the final velocity 12,000 yards/s. The distance over which the capsule accelerates is 700 ft.

    Remember, this is fiction...
  4. Jan 23, 2015 #3


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    Staff: Mentor

    initial speed before the charge is lit is zero

    final speed you know
    distance over which it accelerates is given

    find acceleration (the assumption is that acceleration is uniform)
  5. Jan 23, 2015 #4
    Haha, yes I picked up on that. They would have the speed to get past LEO after 213m.

    So are you saying the final velocity is 12,000 yards/s AFTER the 213m uniformed acceleration?
  6. Jan 23, 2015 #5


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    Staff: Mentor

    Yes, of course. That's how canon's work. The acceleration takes place while the projectile is within the barrel.
  7. Jan 23, 2015 #6
    Ok, I still can't seem to figure out what i'm doing wrong. Here's my work:

    [itex]v^2 = v_0^2 + 2a\Delta x[/itex] Solving for [itex]a[/itex] gives me:

    [itex]\frac{v^2 - v_0^2}{2\Delta x} = a[/itex] Plug in my values:

    [itex]\frac{(12,000 yds/s)^2}{2(233.34 yds)} = a[/itex] Simplify:

    [itex]\frac{144,000 yds/s}{466.67 yds} = a[/itex]

    [itex]308.57 s = a[/itex]

    Some how I end up with only seconds as my unit. I assume I'm using the wrong formula?

    Edit: Ahhhhh, nevermind. I forgot to square my units in the numerator!
  8. Jan 23, 2015 #7


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    Staff: Mentor

    ##\left(\frac{yd}{s}\right)^2 = \frac{yd^2}{s^2}## right?

    And check the order of magnitude for your yards squared. You've lost several.
  9. Jan 23, 2015 #8
    Yes, I just caught that, thank you
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