# Calculate acceleration of a ship fired from a cannon

## Homework Statement

Jules Verne [/B]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

## Homework Equations

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

## 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 $-9.8\frac{m}{s^2}$?

Last edited:

## Answers and Replies

gneill
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...

NascentOxygen
Staff Emeritus
Science Advisor
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)

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...

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?

gneill
Mentor
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?
Yes, of course. That's how canon's work. The acceleration takes place while the projectile is within the barrel.

Ok, I still can't seem to figure out what i'm doing wrong. Here's my work:

$v^2 = v_0^2 + 2a\Delta x$ Solving for $a$ gives me:

$\frac{v^2 - v_0^2}{2\Delta x} = a$ Plug in my values:

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

$\frac{144,000 yds/s}{466.67 yds} = a$

$308.57 s = a$

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!

gneill
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.

Yes, I just caught that, thank you