# Fun with baseball and cannon sim. ?

fun with baseball and cannon sim. plz help?

hey all, I was just curious about a few questions.
ok assuming a baseball weighs 5 ounces (and has a mass of .142kg)
how many newtons of force would it take to shoot the baseball 500meters at a 45degree angle assuming the target was level with the the cannon?
how many newtons to shoot it 100meters at the same angle?

And I don't know if my math is right, but it shows in order to get an object to shoot 500 meters at a 45 degree angle, would require it to travel at roughly 140 meters/sec.
i took
X=(vi^2 sin(theta)^2)/-a
500=((sin(45)Velocity)^2(sin45)^2)/9.8
and from there I get that Velocity = 140 m/sec

(If my math is right) how could i find out how how many newtons it would take to achieve a 140 meter/sec force on the .142kg baseball?
and how many newtons of pressure would it hit the target with?
could you please include your work and explain it as you do it.

this isn't homework, just curiosity.
thanks for any input
-James

Thanks for any help!

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cepheid
Staff Emeritus
Gold Member

And I don't know if my math is right, but it shows in order to get an object to shoot 500 meters at a 45 degree angle, would require it to travel at roughly 140 meters/sec.
i took
X=(vi^2 sin(theta)^2)/-a
500=((sin(45)Velocity)^2(sin45)^2)/9.8
and from there I get that Velocity = 140 m/sec
I don't think your formula for the range of the projectile is quite right. It should be:

$$\Delta x = \frac{2v_{0}^2\cos{\theta}\sin{\theta}}{g}$$

$$= \frac{v_{0}^2\sin{(2\theta)}}{g}$$​

You derive this by finding the time to reach the max height, doubling it to get the total travel time, and then multiplying this by the horizontal component of the initial velocity.

cepheid
Staff Emeritus
Gold Member

how could i find out how how many newtons it would take to achieve a 140 meter/sec force on the .142kg baseball?
and how many newtons of pressure would it hit the target with?
Pressure is NOT measured in newtons. Force is measured in newtons. I am not trying to nitpick: force and pressure are two different physical quantities (albeit closely related to each other) with two different definitions. If it is the force you are looking for, then here's the thing: you can use any amount of force you want to achieve that final speed. The only thing that has to change is the amount of time over which the force is applied. You can use a small amount of force and apply it over a long duration, or use a larger force applied over a smaller duration. To see why this is true, start with Newton's second law: the force on the object is equal to its mass times its acceleration:

$$F = ma$$​

If we assume the force is constant, then the acceleration is constant, and hence it is just equal to the total change in velocity divided by the time over which the change occurred.

$$F = m\frac{\Delta v}{\Delta t}$$​

Now, the product of the mass and the change in velocity is just the change in momentum of the object:

$$F = \frac{\Delta p}{\Delta t}$$

$$F\Delta t = \Delta p$$​

The quantity on the left-hand side is called the impulse delivered to the object. This equation is an expression of the impulse-momentum theorem. The impulse delivered to the object is equal to its change in momentum. Your condition requires that the ball gain a certain amount of momentum. In other words, it requires that a certain amount of impulse is provided by the cannon. Unfortunately, that condition alone doesn't constrain how much force should be applied, since the impulse is the product of the force and the time interval.

In a cannon, the amount of time over which the force can be applied is limited by the duration spent in the barrel. Unfortunately, you have no idea how long this is. Another problem is that it's probably not very realistic to assume that the force provided by the cannon will be constant. However, even if the force is changing with time, my last equation is still true, provided you replace F with the average force (the average being over the time interval in question).

EDIT: A simpler way to understand this conclusion (directly from Newton's 2nd): the force applied determines the acceleration, which is the rate of change of the velocity. So, applying a large force will cause the velocity to change rapidly, and it will take less time to reach the desired launch speed than a smaller force, which would cause a smaller acceleration and hence require more time to reach the desired final speed at launch.

ok thanks for the input.
I obviously have a lot to learn about physics.

But is it even realistically possible to make a spring-canon (a cannon with an industrial spring and wrenching system) to shoot something like a baseball over large distance (like a 100 meters).

I guess what I am trying to figure out is what kind of a spring I would need, how long of a tube, and what kind of ammo would be best use to make a real life spring cannon for a science project.

Would it be possible to do something like this at a home-made level?
And what further online reading material would you suggest for similar topics?

Thanks again!
-James

cepheid
Staff Emeritus
Gold Member

Well, that might be feasible! At first, I was thinking you were talking about a cannon that ignites gun powder and propels a projectile using the rapidly expanding gases from the explosion. That is something that would be very difficult to model theoretically.

For a spring, the major parameter that you would need to know is the spring constant. A system consisting of a mass attached to the end of a spring is often modelled using Hooke's law, which says that the spring provides a restoring force that is always in opposition to the displacement of the spring:

F = -k(x-x0)

where x0 is the equilibrium position of the spring (i.e. the position of the mass for which the spring is uncompressed, and hence the restoring force is 0). It's called the restoring force because, as you can see, its sign is always opposite to the sign of (x-x0). If you compress the spring (meaning x < x0 and hence (x-x0) < 0), the force is positive (meaning it pushes back in the direction that makes the spring uncompressed). If you pull on the spring to try to expand it, (meaning x > x0 and hence (x-x0) > 0), the force is negative, meaning that it tries to pull back on the mass in the direction of spring compression. So, either way, the force tries to restore the spring to its equilibrium position (neither compressed nor expanded).

Hooke's law is not really a law in the sense that a spring doesn't *have to* obey it, but this turns out to be a good model for many springs. The spring constant k (measured in N/m), varies from one spring to another. It is a parameter that characterizes the "stiffness" of the spring, because it tells you how much restoring force you are going to get PER metre of displacement.

The nice thing about the spring force is that it is conservative, meaning that it is associated with a potential energy function. What I mean by conservative is that any kinetic energy that the mass loses due to compression or expansion of the spring is stored in that spring as *elastic potential energy* (and can be recovered later). The expression for the elastic potential energy for a "Hookian" spring is:

E = (1/2)k(x-x0)2

By the way, for potential energy, only changes matter. In other words, you can define the equilibrium position to be whatever you want. For instance, you could put it at the origin of your coordinate system so that all other displacements would be measured relative to it. In other words, you could set x0 = 0, thus simplifying the equations.

So, using this last equation, and assuming that all of the elastic potential energy stored in the spring when it is depressed will be converted into kinetic energy once it is released (i.e. at launch), you can figure how FAR you need to depress the spring by setting the potential and kinetic energy equal to each other:

E = (1/2)kx2 = (1/2)mv02

This solves your problem. All you have to do is measure m, measure k for the specific spring that you have bought, and then determine the value of x according to the desired launch speed v0. Once you know x, you know the force with which the spring must be held compressed (before it is released for launch).

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