# Rifle shot

1. Feb 3, 2008

### AnkhUNC

[SOLVED] Rifle shot

1. The problem statement, all variables and given/known data
A rifle that shoots bullets at 475 m/s is to be aimed at a target 42.5 m away. If the center of the target is level with the rifle, how high (in cm) above the target must the rifle barrel be pointed so that the bullet hits dead center?

2. Relevant equations

sin(2theta) = gd/v0^2

3. The attempt at a solution

So sin(2theta) = (9.8)*(4250)/(47500^2) = 1.84598338e-5

So sin(2theta) = ?

2. Feb 3, 2008

### timon

first calculate how long it would take the bullet to reach the target

then using that time and the earths acceleration, calculate how much it must have dropped at the end

this is the height b

a is the horizontal distance
http://em-ntserver.unl.edu/Math/mathweb/trigonom/Image289.gif [Broken]

calculate the angle

i'm not to sure you can use a right triangle here though, so don't take my word for it :D

Last edited by a moderator: May 3, 2017
3. Feb 3, 2008

### Kurdt

Staff Emeritus
One needs to work out in cm how high the barrell must be pointed. That means that the gun will be parallel to the ground and all you have to do is work out how far the bullet will deviate vertically due to gravity.

4. Feb 3, 2008

### AnkhUNC

So I have v0, x0 and xF and G. I need to find t in order to find y0 right? I can just assume that the ground is parallel to yF would be 0?

5. Feb 3, 2008

### Kurdt

Staff Emeritus
Yes you will need to find t. Just assume the middle of the target is the 0 point for height.

6. Feb 3, 2008

### AnkhUNC

Is there an equation that I don't have to know the angle in order to solve?

7. Feb 3, 2008

### Kurdt

Staff Emeritus
Oh sorry I think I mis-read the question. It is asking for the angle I think, but to me its not written very clearly.

8. Feb 3, 2008

### AnkhUNC

So my original answer is right excluding the fact that its wrong?

9. Feb 3, 2008

### Kurdt

Staff Emeritus
Your original approach is correct except you seem to be multiplying some of the numbers by 100 for no reason. Once you have the angle you can use trig to find the height above the target centre.

10. Feb 3, 2008

### AnkhUNC

I was doing that so I wouldn't have to convert to cm later on. sin(2theta) = 1.84598338e-5 is right? So how do I just find the angle?

11. Feb 3, 2008

### Kurdt

Staff Emeritus
Keep it in meters and convert later. Its not correct because the acceleration due to gravity is in ms-2. Once you have $\sin(2\theta) = ....$ you need to take the inverse sin of the number you get. That will give you twice the angle.

12. Feb 3, 2008

### AnkhUNC

So (-9.8 m/s^2)(42.5)/(475^2m/s) = sin(2theta)? = -.0018459834

Theta = -1.057671168? But that doesn't seem right.

13. Feb 3, 2008

### Kurdt

Staff Emeritus
You're a decimal place out and you can lose the minus sign. The angle you have there is twice the angle you want. Remember the bullet travels very fast so the angle will only be small.

14. Feb 3, 2008

### AnkhUNC

So .0018459834 should be .00018459834? /2 = the angle I want?

15. Feb 3, 2008

### Kurdt

Staff Emeritus
$\theta$ = 0.106/2. Then one can work out how far above the target the bullet will be aimed through trigonometry.

16. Feb 3, 2008

### AnkhUNC

42.5*tan(.053) = .393135526m = 3.93cm! Thanks!

17. Feb 3, 2008

### Kurdt

Staff Emeritus
Sorry for the initial confusion.