# Civil War Cannon physics

1. Dec 30, 2013

### Medgirl314

1. The problem statement, all variables and given/known data
A Civil War cannon fires a cannonball at 900 ft/s. If the cannon is aimed 5 degrees above the horizontal, find the time that the cannonball is in the air.

A Civil War cannon fires a cannonball at 900 ft/s. If the cannon is aimed 5 degrees above the horizontal, find the height that the cannonball reaches.

2. Relevant equations
Sin=opp/hyp
y=y0+v0t+1/2a(t^2)

3. The attempt at a solution

First I found the vertical velocity using sine.
900 sin s=78 ft/s .

Next I found the time using the vertical velocity.

v0=78 ft/s
a=-32.2 ft/s
v=0
t=(v-v0)/a
0-78/-32.2
t=-78/-32.2
t=4.8 s.

4.8 seconds should be the answer to the first problem.

Next, I used y=y0+v0t+1/2a(t^2) to find the height the cannonball reached.

y=0+78(4.8)+1/2(-32.2)(4.8)
y=374.4+1/2(154.56)
y=377.4-77.28=300.12

Height=300 ft.

Are my answers correct? Thanks in advance!

2. Dec 30, 2013

### SteamKing

Staff Emeritus
You need to check your calculation of 78/32.2

3. Dec 30, 2013

### Medgirl314

Ah, I see. I doubled it for some reason. t=2.4 s.

4. Dec 31, 2013

### Medgirl314

Oh, I doubled it because I considered the motion from the start to the peak. So to find the total time, I doubled my start to peak time.

5. Jan 4, 2014

### Medgirl314

Could someone please either verify or correct my calculation of the time?

6. Jan 4, 2014

### haruspex

When you evaluate some intermediate quantity (vertical speed in this case) then plug that number into another equation to get the quantity you want, you run the risk of error accumulation. You need to maintain more significant figures than you will quote in the final answer.
Also, yes, you correctly doubled the time to get time in the air, but then you used that doubled time to find the height. Reconsider that.

7. Jan 4, 2014

### Medgirl314

Okay. You're saying to re-work the vertical speed just like I did, but maintaining more decimal places, and then using the original(not doubled) speed to find the height. Correct?

Thank you!

8. Jan 4, 2014

Yes.

9. Jan 4, 2014

### Medgirl314

I'm sorry, but I'm having a hard time seeing how I found v0 using sine the first time. I thought I had posted it clearly, but it looks like I may have made a typo. Do you know how I did it the first time?

Thanks again!

10. Jan 5, 2014

### haruspex

You mean, 900 sin(5 degrees) ft/sec?

11. Jan 5, 2014

### Medgirl314

Yes, thank you! I accidentally left out the (5 degrees), which, for some reason, messed with my brain.

v0=78.44 ft/s
a=-32.2 ft/s
v=0
t=(v-v0)/a
0-78.44/-32.2
t=-78.44/-32.2
t=2.44 s.

Total time, and final answer for the first problem: 4.9 s

(Copying and editing my OP)

Next, I used y=y0+v0t+1/2a(t^2) to find the height the cannonball reached.

y=0+78.44(2.436)+1/2(-32.2)(2.436)
y=191.07+1/2(-78.439)
y=191.0-39.32 = approximately 152 ft.

Thanks so much for the tips! How does that look?

12. Jan 5, 2014

### Tanya Sharma

You missed squaring the term marked in red.

13. Jan 5, 2014

### Medgirl314

Yes, I did miss that. Thank you! Once I learn LaTex seeing how to solve the equations will be earlier. I am on mobile and new to the app,so I can't go back and forth between my last post and this one. Look for my answer in about 10 hours, if you don't mind, as it is time for bed here!

Sent from my iPhone using Physics Forums

14. Jan 5, 2014

### Medgirl314

y=0+78.44(2.436)+1/2(-32.2)(2.436)
y=191.07+1/2(-32.2)(5.934)
y=191.07-95.53
y= approximately 96 feet.

Thanks again, Tanya Sharma and haruspex!

How does that look?

15. Jan 5, 2014

### haruspex

Yes.

16. Jan 5, 2014

### Medgirl314

Great, thank you! That should be it for these questions, yes?

17. Jan 5, 2014

### Medgirl314

When you have a moment, could you please check the other problem we were working on?