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

by jsewell
Tags: angle, distance, formula, sniper, target
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jsewell
#1
Feb26-12, 08:29 PM
P: 10
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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Rap
#2
Feb26-12, 09:18 PM
P: 789
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 thats not what you want.
genericusrnme
#3
Feb26-12, 09:35 PM
P: 615
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?

jsewell
#4
Feb26-12, 09:37 PM
P: 10
Formula for the angle a sniper must make to hit a target at distance x

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?
Rap
#5
Feb27-12, 12:08 AM
P: 789
Quote Quote by jsewell View Post
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, thats 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.


Quote Quote by genericusrnme View Post
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
willem2
#6
Feb27-12, 12:43 AM
P: 1,395
Quote Quote by Rap View Post
But if you say that y=y0, thats 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]
jsewell
#7
Feb27-12, 01:04 AM
P: 10
How might you have it willem2?
willem2
#8
Feb27-12, 01:10 AM
P: 1,395
Quote Quote by jsewell View Post
How might you have it willem2?
I get

[tex] x = \frac {F^2 sin {2 \theta}} {g} [/tex]
jsewell
#9
Feb27-12, 04:19 AM
P: 10
Sooo the formula for the angle needed is? I don't mean for this to sound rude.

I solved another problem in my text book 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 cancelled perfectly. Could this truely a working formula?
Rap
#10
Feb27-12, 07:30 AM
P: 789
Quote Quote by willem2 View Post
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.
jsewell
#11
Feb27-12, 12:13 PM
P: 10
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


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