Initial velocity out of a cannon

[flamedfordbronco]

Hey everyone,
I'm doing a lab for engineering class, and I'm having a little trouble with a part of it. The lab was that we have a cannon on a table, with 3 launch settings (different lengths the spring is compressed). The table is 40" high. Also, we have a dixie cup 90" away from the table that we have to shoot the ball into. The problem was to find the angle to put the cannon at so that the ball lands in the dixie cup.
The problem I'm having is with finding the initial velocity of the ball out of the cannon. The spring constant is 235 N/m, and the amount the spring is compressed is 0.123825 m. The ball weighs 67 g. Now, we did the experiment in class, and the angle was around 61 degrees (we just found out with trial and error), and I think i can figure it out with calculations once I get the initial velocity, but how do I find that? Thanks a lot in advance for your help.

~Frank~

 

[Integral]

I do not have the relationships in my head, but you should be able to find the Potential energy of the spring, use this to get the KE and and then the velocity.

 

[JasonRox]

Find out the time it remains in the air.

Shoot it straight up three times, and time it.

Using Calculus, or formula's from a Physics book, you should find a good approximation of the initial velocity.

As far as I am concerned, the initial velocity does not change significantly when you change the angle. (0,45, or 90)

 

[flamedfordbronco]

Integral,
Yeah our professor did something with the PE and KE, but I couldn't follow him. I copied down what he wrote on the board but i just don't understand it. Here's what he wrote- PE=1/2 kx^2
KE= PE at x=0
KE=1/2 mv^2
PE=1/2(235N/m)(.0635 m)^2
PE=0.48 Nm
V=(0.48N m)^2/(0.025)
V=6.2 m/s
.025 is the example weight he used for the projectile

If you understand all this stuff, could you explain it to me? Thanks a lot!

Jason,
yeah that's exactly what our group thought of doing, but we didn't do the lab on our own. our professor did it in front of the class, and we all had to tell him what angle to shoot it at, using the numbers he gave us. Thanks to both you guys for the posts!

~Frank~

 

[Integral]

Ok, those are the equations we need.

You have the PE of the spring as

$$PE= \frac 1 2 k x^2$$

You know that PE = KE so

$$PE=KE = \frac 1 2 m v^2$$

so

$$v = \sqrt \frac {2\ PE} m$$

Where the m is the mass of your ball.

Now we can plug in the expression for PE to get

$$v = \sqrt {\frac k m } x$$

The final check will be to verify the units of the answer.
k is in N/m and mass is in kg. N = kg m/s^2 so we have
$$\sqrt {\frac {kg \frac m {s^2}m} {kg}$$
now cancel the kg in the numerator and denomiator and combin the meters to get
$$v= \sqrt {\frac {m^2} {s^2}} = \frac m s$$

Our result has units of velocity so we can be sure that it has some meaning.