# Need Help: Momentum and Energy

1. Dec 5, 2009

### Edgar92

1. The problem statement, all variables and given/known data
A Hunter shoots a 500g arrow at a 2 kg bird perched on a tall tree growing on flat, level ground. The arrow is launched from ground level with a speed of 40 m/s at an angle of 30 degrees above the horizon. It is travelling horizontally when it embeds in the bird. How far from the base of the tree do the bird and aroow land

2. Relevant equations
Momentum and Energy Question
P=m*v
Ek=m*v*v/2
Eg=m*g*h
Etot1=Etot2
P1=P2

3. The attempt at a solution
V=40m/s*cos30 = 34.64
P1=.5kg*34.64m/s= 17.32
P2=V2*2.5kg
17.32/2.5=6.93=V2
.5*6.93*6.93/2+.5*9.8*h=????

I have been supplied with the correct answer, but I must figure out how to get there (14m)

2. Dec 5, 2009

### denverdoc

Couple of different ways to tackle this. First you need h, and this can be easily solved by looking at the vertical component of the initial velocity, i.e. sin (30)* Vi. Once known, treat is as a simple ballistics problem with the new velocity you obtained by conserving mo.

3. Dec 5, 2009

### Edgar92

20 m in height
Alright Ill try it again and if it diesn't work Ill post in like 5 minutes

4. Dec 5, 2009

### Edgar92

So i just tryed it again and got 17.8 so Im thinking one of my numbers I provided in my attempt posted above must be wrong

5. Dec 5, 2009

### denverdoc

What t did you get?

6. Dec 5, 2009

### Edgar92

I never got a t because Ek is mv^2 and Eg is mgh
I didn't know t was relevant nor have I thought about how to get it from my givens

7. Dec 5, 2009

### denverdoc

Ok, as I said, there is more than one way to approach the problem. I verified the answer as 14 using t.

8. Dec 5, 2009

### denverdoc

Maybe if you show me the eqn you used to get the answer 17 I can help.

9. Dec 5, 2009

### Edgar92

I'll post in a sec

10. Dec 5, 2009

### Edgar92

So i got Vi with 40cos30 which gave me 34.64m/s
i used momentum p1=p2 and did the following
.5(arrow)*34.64=2.5(arrow+bird)*Vf
I got 6.93m/s
from there I did Ek(arrow)+Eg(bird)=Ek(final)
.5*34.64^2/2+2*9.8*20=2.5*Vf^2/2
300+392=2.5Vf^2/2
692*2=2.5Vf^2
1384/2.5=Vf^2
553.6^(1/2)=Vf
I tried using the bird for Eg but obviously that worked no better as my number came out as 23.5 but that is what i tried

11. Dec 5, 2009

### Edgar92

Ill be at work for the next 7 hours so I will be restricted to the work I can do but I will still check this site on my Iphone and can try stuff while it is slow at work

12. Dec 5, 2009

### denverdoc

But I still don't see where this leads you to x, the horizontal distance from the tree? The math looks good, what I did was to solve for t from the initial vertical velocity--you don't even need to compute h, but thats fine if you do. Here t is the time of ascent which is also the time of descent. You have computed the horizontal velocity after impact, you have t, the answer awaits.

13. Dec 5, 2009

### Edgar92

I know t?

14. Dec 6, 2009

### vu95112

Need help on Plane Mirror

Plane Mirror
A person walks into a room that has, on opposite walls, two plane mirrors producing multiple images. Find the distances from the person to the first three images seen in the left-hand mirror when the person is 5ft from the mirror on the left wall and 10 ft from the mirror on the right wall.

I know the first image is 10 ft because in the plane mirror, the image is as far behind the mirror as the object is in front. I don’t know how to figure out the second and the third images? I need you help please. Thank you.

Vu95112

15. Dec 6, 2009

### Edgar92

yo maybe we can finish my question first and then move on to yours

16. Dec 6, 2009

### denverdoc

Sure you can find t. The most straightforward is to get Vy from the initial data and divide by g. Say Vy=30m/s then t is about 3 seconds.

17. Dec 6, 2009

### Edgar92

Thanks man, you been a great help. I was stressing out about this before, but you really saved my @

18. Dec 6, 2009

### denverdoc

No problem.

19. Dec 6, 2009

### Edgar92

Sorry to post again but I read and made that last post at work.
I am assuming you mean g is gravity(9.8) but that gives me 3.53 s. am I just supposed to use d=v*t to solve.

20. Dec 6, 2009

### denverdoc

Thats not right--the time is (sin 30* 40m/s)/9.8 should be about 2 seconds. In other words the vertical velocity is 20m/s.