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

jsewell

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

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

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

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

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.


You wrote it wrong, a_b_c is not allowed, maybe a_{b_c} or {a_b}_c

willem2

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

How might you have it willem2?

willem2

I get

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

jsewell

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

Ok, I see, the t is the t when the bullet hits the aim point.

jsewell

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