Formula for the angle a sniper must make to hit a target at distance x

In summary, a person was trying to determine the angle a sniper must make with his barrel to hit a target over a certain distance, neglecting external factors. They used the standard equations of motion and made a mistake by assuming the y-coordinate stayed constant. They also made a mistake in their equation for the angle needed. After some revisions and using Firefox as their browser, they were able to reach a working formula for the angle.
  • #1
jsewell
10
0
I just got curious today and tried to determine the angle a sniper must make with his barrel to hit a target over some distance, neglecting air resistance, humidity, etc. Any suggestions as to why my solution on the bottom right is wrong?
 

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  • #2
You have y=yo+vo t - g t^2/2 and then you have 0 = vo t - g t^2/2 so that means y=yo always, and that's not what you want.
 
  • #3
Your handwriting is neat but your image is on it's side :(

The standard equations of motion for an object in a uniform gravitational field is

[itex]y=y_0 + v_y_0 t -\frac{1}{2} g\ t^2[/itex]
[itex]x=x_0+v_x_0 t[/itex]

Edit;
LaTeX doesn't work on these boards?
 
  • #4
Yes, I did this to solve for T. Afterwords I multiplied that formula by 2 in order to determine the total air time. Once I found the total air time formula T= 2Fsin(θ)/2 I plugged it into the formula X = X° + Vxo * T. Was this an incorrect strategy?
 
  • #5
jsewell said:
Yes, I did this to solve for T. Afterwords I multiplied that formula by 2 in order to determine the total air time. Once I found the total air time formula T= 2Fsin(θ)/2 I plugged it into the formula X = X° + Vxo * T. Was this an incorrect strategy?

But if you say that y=y0, that's wrong. You are saying that the y coordinate never changes, always stays where it started from. If you make that wrong assumption, you are bound to get wrong results.


genericusrnme said:
Your handwriting is neat but your image is on it's side :(

The standard equations of motion for an object in a uniform gravitational field is

[itex]y=y_0 + v_y_0 t -\frac{1}{2} g\ t^2[/itex]
[itex]x=x_0+v_x_0 t[/itex]

Edit;
LaTeX doesn't work on these boards?

You wrote it wrong, a_b_c is not allowed, maybe a_{b_c} or {a_b}_c
 
  • #6
Rap said:
But if you say that y=y0, that's wrong. You are saying that the y coordinate never changes, always stays where it started from. If you make that wrong assumption, you are bound to get wrong results.

He substituded 0 for y_0, because y_0 happens to be 0, and then set the remaining expression for y equal to 0. There's nothing wrong with that

However, there's an error (or even 2 errors) between

[tex] x = \frac { 2 F \cos {\theta})F \sin {\theta} } {g} [/tex]

and

[tex] x = \frac {F \sin {2 \theta} } {g} [/tex]
 
  • #7
How might you have it willem2?
 
  • #8
jsewell said:
How might you have it willem2?

I get

[tex] x = \frac {F^2 sin {2 \theta}} {g} [/tex]
 
  • #9
Sooo the formula for the angle needed is? I don't mean for this to sound rude.

I solved another problem in my textbook very similar to this one where a fire hose shot water at a velocity of 6.5 m/s. I had to determine was angle(s) the nozzle could be at to reach a target 2.5 meters away. I worked the problem very similarly to my OP and found the angle of 18* to work. The book noted a second angle at 72*. I suppose 90 - 18 = 72 so I can somewhat visualize this second result though still fail and finding it personally. The equation I used was the same for my sniper question...

(1/2)arcsin(xg/v^2) = theta

Where x is displacement in meters
G is the gravitational constant in m/s^2
V is the velocity in m/s

I also did a dimensional analysis on the argument of arcsin and found all the units canceled perfectly. Could this truly a working formula?
 
  • #10
willem2 said:
He substituded 0 for y_0, because y_0 happens to be 0, and then set the remaining expression for y equal to 0. There's nothing wrong with that

However, there's an error (or even 2 errors) between

[tex] x = \frac { 2 F \cos {\theta})F \sin {\theta} } {g} [/tex]

and

[tex] x = \frac {F \sin {2 \theta} } {g} [/tex]

Ok, I see, the t is the t when the bullet hits the aim point.
 
  • #11
lol just used Firefox as my browser and willem2s formula looks far more comprehendable

Yes willem2 I did reach that formula after some revisions. I didn't square the Velocity of the projectile (bullet). Although after solving the same equation for theta are you able to reach my formula of

(1/2)(arcsin(XG/V^2) = theta

? Idk the fancy coding needed to make this formula look neat yet hah
 

1. What is the formula for calculating the angle a sniper must make to hit a target at distance x?

The formula for calculating the angle a sniper must make to hit a target at distance x is: angle a = arctan (height difference / distance).

2. How do you determine the height difference in the formula?

The height difference is the difference in height between the sniper's position and the target's position. This can be determined by measuring the vertical distance between the two points.

3. What unit of measurement should be used for distance in the formula?

The unit of measurement for distance in the formula should be consistent with the unit of measurement used for the height difference. For example, if the height difference is measured in meters, then the distance should also be measured in meters.

4. Is the formula affected by other factors such as wind or elevation?

Yes, the formula for calculating the angle a sniper must make to hit a target at distance x is affected by other factors such as wind and elevation. These factors can change the trajectory of the bullet and should be taken into consideration when using the formula.

5. What is the purpose of using this formula for sniping?

The purpose of using this formula is to help snipers accurately hit targets at long distances. By calculating the angle a sniper must make, they can adjust their aim and account for any height differences between them and their target, increasing their chances of hitting the target with precision.

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