# Homework Help: 2 dimensional problem with only angle and direction.

1. Oct 8, 2011

### freeofwork

1. The problem statement, all variables and given/known data
A cannon fires a cannonball 500.0 mdownrange when set at a 45 degree angle. At what velocity does the cannonball leave the cannon?

2. Relevant equations

Vix=Vi X cosθ
Viy=Vi X sinθ
Tair=viy/a X 2
D=1/2 X at2
D=vt

3. The attempt at a solution

If i decide to use d=1/2 X at2 will the acceleration be 0? or 9.81? Because isnt the acceleration 0 here because the x component has no acceleration? But that makes no sense at all because the result would have to be 500m. And 9.81 is the y component acceleration. So i not sure.

2. Oct 8, 2011

### Rayquesto

What does "down range" mean? does it imply that the angle that the cannon shots at is downward?

3. Oct 8, 2011

### freeofwork

Im guessing it means that the distance traveled from one point to another is 500m

4. Oct 8, 2011

### Rayquesto

if that were the case, it would make the prolem very hard to solve. This would be like a UC berkeley type problem. I already tried solving it that way, but there's something I'm missing. Something I can't prove right, but maybe I'm miss reading what this means. So, what does "down range" imply?

5. Oct 8, 2011

### Rayquesto

ok, well if thats what it means, then it's mUUUUCh easier to solve.

from this equation: Vfy=Viy - gt

in what situation do you think the motion makes Vfy=0m/s?

6. Oct 8, 2011

### freeofwork

The thing i am not quite getting is that how is the acceleration 9.81 m/s2. Isnt 500 m only the horizontal direction. Because the parabola created by the initial elocity frm the degree measure would actually have traveled more?

7. Oct 8, 2011

### Rayquesto

acceleration in the y direction is always -9.81m/s^2 as long as it is in the air, but the acceleration in the x direction is zero. hold on.

8. Oct 8, 2011

### Rayquesto

500meters is only the horizontal range. that still means theres a height that they dont give you though.

9. Oct 8, 2011

### freeofwork

Can anyone else confirm?

10. Oct 9, 2011

### Rayquesto

think about it. If you launch something at 45 degrees above horizontal, then won't there be a change in height and a change in range? range is the horizontal distance.

11. Oct 9, 2011

### freeofwork

So there is a piece missing correct?

12. Oct 9, 2011

### Rayquesto

you wont need height to find out what they are asking.

I'll give you the formula you need to use, but I'd like that you figure out how it was derived. I'll help you too. :)

Range=V0^2Sin(2theta)/g

13. Oct 9, 2011

### Rayquesto

SO, as I was saying

from this equation: Vfy=Viy - gt

in what situation do you think the motion makes Vfy=0m/s?

14. Oct 9, 2011

### freeofwork

I have no idea...

15. Oct 9, 2011

### Rayquesto

it's when the maximum height occurs, and when half the maximum range occurs.

so, if t=Viy/g at that point, then 2Viy/g=t for the entired motion to take place and when range exists.

And then distance=Vix(t)

so, (range)=vix(2viy/g)

vix=vi(costheta)
viy=vi(sinetheta)

range=2vi^2costhetasintheta/g

range=vi^2sin2theta/g