
#1
Feb2612, 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?




#2
Feb2612, 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.




#3
Feb2612, 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? 



#4
Feb2612, 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?




#5
Feb2712, 12:08 AM

P: 789





#6
Feb2712, 12:43 AM

P: 1,351

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
Feb2712, 01:04 AM

P: 10

How might you have it willem2?




#8
Feb2712, 01:10 AM

P: 1,351

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



#9
Feb2712, 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? 



#10
Feb2712, 07:30 AM

P: 789





#11
Feb2712, 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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