Can any Maths/Physics gurus help please?

Si

I need to know the Force that a cannon exerts on a cannonball, ie. The force the cannonball travels at.. I know the velocity of the ball as it leaves the cannon (1159mph), and I know the weight of the cannonball (12 pounds). I also know the F=MA equation, but am unsure as to how velocity and acceleration can be interchanged..

If anyone can help in a timely fashion I would be VERY grateful indeedy

Ta all...

WSS

Lol

KillersBeard

Aren't they equal?

Something to do with Newtons law and equal and opposite.

But I could be talking bollocks??????????

Elthamaddick

whooooosh......pass, sorry!

WSS

Kanu?

JT

Better of posting this on Millwall online.
They're better at things like this.

KillersBeard

They would certainly be able to help with knuckle gravity!

Stig

To work out the acceleration you need to now the time taken from a standing start to reach the given speed. It is usually measured in metres per second per second.

ValleyGary

F=U+C/K (KxN+0=WS)

Cordoban Addick

[cite]Posted By: ValleyGary[/cite]F=U+C/K (KxN+0=WS)

Excellent, pissed myself laughing.

razil

If you are studying Physics then surely you have a text book to tell you this, if not perhaps you should be..

Si

I'm actually at work - this is the sort of thing my day is filled with.

Thanks for all your posts - Google has finally come to the rescue, as always.

For those who are interested:

http://waowen.screaming.net/revision/force&motion/ncananim.htm

EDIT: Actually, as good as this looks, it doesn't take into account the weight of my balls, so to speak. So no good.

3blokes

At last a post about this :-)
It's been bothering me for weeks ;-)

Dave Rudd

Ahem ...

.. it goes like this:


If you know the velocity of the cannonball, it is easy to calculate the acceleration using the following formula:

v squared = u squared + 2as

where v is the final velocity, u is the initial velocity ( = 0 in this example), s is the distance travelled (so the length of the cannon) and a is the acceleration. Once you have the acceleration, your F = ma equation can come into play.

So, the main info which seems to be missing is the length of the weapon ( ooer!). No exaggeration, please.

Also, make sure you get your units right.

Now bring on Stephen Fry.

KillersBeard

[cite]Posted By: Dave Rudd[/cite]

So, the main info which seems to be missing is the length of the weapon ( ooer!). No exaggeration, please.

And of course the weight of his balls!!!!!!

Has this stopped being a physics question now???

incorruptible addick

What are you going to do with these cannon balls ?

If you've nothing else planned, can I suggest firing them up the backsides of our team, perhaps?

KillersBeard

Just don't ask any of our lot to aim and fire, or we'd need a new stand!

Stig

[cite]Posted By: Si[/cite]I'm actually at work - this is the sort of thing my day is filled with.

You're not a supply teacher are you?

Friend Or Defoe

I'm not going to do it for you, but you work out the angle of the cannon you can use trigonometry to calculate it's projection, add in gravity at 9.8m per a second per a second and you have a formula for it's speed, differentiate this for the acceleration.

There should be loads of examples on google/your text book.

C.Walsh'sLoveChild

If you was to hold a loaded gun in one hand and a bullet in the other at chest height. Then at the same time fire the gun and drop the bullet. Which bullet would hit the ground first? Providing obviously that the gun was exactly horizontal to the ground

Henry Irving

[cite]Posted By: C.Walsh'sLoveChild[/cite]If you was to hold a loaded gun in one hand and a bullet in the other at chest height. Then at the same time fire the gun and drop the bullet. Which bullet would hit the ground first? Providing obviously that the gun was exactly horizontal to the ground

I saw QI last week too

dabos

They would hit the ground at exactly the same time.

aliwibble

Theoretically yes, in reality no, as they've ignored the effect of air resistance on the bullet. Admittedly it'll be a lot less than if you carried out the same experiment with a brick or other not very aerodynamic thing, but they won't land at exactly the same time.

dabos

[cite]Posted By: aliwibble[/cite]Theoretically yes, in reality no, as they've ignored the effect of air resistance on the bullet. Admittedly it'll be a lot less than if you carried out the same experiment with a brick or other not very aerodynamic thing, but they won't land at exactly the same time.

My answer was a theoretical one rather than an 'in reality' one

Henry Irving

You'd also have to allow for the curvature of the earth : - )

C.Walsh'sLoveChild

I thought i might of been able to pass that off as a genuine piece of knowledge that i had gained. My QI source has been rumbled

Dave Rudd

[cite]Posted By: Henry Irving[/cite]You'd also have to allow for the curvature of the earth : - )

... and the type of gun, the shape, size and weight of the bullet, the height above sea level, the nature of the terrain, the wind speed and direction, the proximity of any buildings or people etc.

Anyone see the Horizon programme on Infinity last week? It went on a bit.

Henry Irving

[cite]Posted By: Dave Rudd[/cite]

Anyone see the Horizon programme on Infinity last week? It went on a bit.

No, but I watched the one of gravity. It got me down

stonemuse

[cite]Anyone see the Horizon programme on Infinity last week? It went on a bit.

Infinity - Guru Josh ...great song.

Dave Rudd

[cite]Posted By: Henry Irving[/cite]

[cite]Posted By: Dave Rudd[/cite]
Anyone see the Horizon programme on Infinity last week? It went on a bit.

No, but I watched the one of gravity. It got me down

I missed the one on Time Travel .....



*the open goal yawns*

somerton

Use the formular a=v-u/t where a is acceleration, v is final velocity in meters per second, u is the initial velocity in meters per second, and t is the time in seconds for the object to go from initial to final velocity. This formular is only good for a uniform rate of change from u to v.