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Velocity and Sound to determine distance 
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#1
Jun2110, 04:21 PM

P: 8

I have a question that I'm not able to figure out by myself.
I shoot a rifle at a steel target. The projectile, when leaving the muzzle, is traveling at 2650 feet per second (fps). At 200 yards, the velocity has slowed to 2208 fps. At 400 yards, the velocity is now at 1811 fps. I record the time that elapses between the rifle being fired and the "gong" that I hear from the projectile impacting the steel target. The time between the shot and the "gong" is 2.74 seconds. How far away is the steel target? Thank you very much for your help. It's driving me crazy trying to figure this out. I think I understand the forum format better and I apologize for not knowing it before I posted. This is an actual situation, not something that is hypothetical. My attempt to figure this out was by the travel of sound only from the gunshot to the "gong." I didn't take into consideration that the bullet traveling to the target was going greater than the speed of sound. I have trouble sleeping sometimes so I get onto youtube and this night I wanted to see if I could mathmatically verify the claimed distance in the shot. Here is the video of the shot. http://www.youtube.com/watch?v=aiyiBDlRLoE I used the second shot because the "gong" from the first shot was not that noticable. If the bullet was traveling at a constant, I don't think it would be hard to figure. But, the velocity of the bullet continues to decrease wtih distance. Here is where I get lost. I arrived at the 2.74 seconds from timing the shot and "gong" six times then taking an average. So, I would like not only the answer, but also to understand how to arrive at the time of impact with a decreasing velocity. Thank you, Mike 


#2
Jun2110, 07:43 PM

P: 153

In the homework section, I think you need to show your math for the time of travel to the target and then for the time it take for the sound of the gong to get back to you before you can get much of an answer.



#3
Jun2110, 08:20 PM

P: 66

because you must take into account air resistance which varies as the square of the speed you require some differential equations i think
the solution i just attempted didn't work at all so i won't bother posting it how do you know speeds at a certain distance anyways given that this is not a hypothetical question? 


#4
Jun2110, 08:33 PM

P: 59

Velocity and Sound to determine distance
Okay, lets say you are shooting at sea level. The time between the exit of the bullet and the time you hear the sound is composed of: t= X/Vb + X/Vs. Where the first part is the time it takes the bullet to reach the steel target; and the second part is the time it takes for the sound to come back to you. X is the distance to target, Vb is the velocity of the bullet and Vs is the speed of sound. Taking Vs=340.3 m/s, all calculations are done in meters; we have:
2.74 = (X/Vb) + (X/340.3), now we have two unknowns and one equation therefore we’ll try to eliminate Vb. In order to do this will find a relationship between velocity and X using the information given. I made a linear approximation and came up with an equation relating the two using the first and last point Vb=2.22X+2650. No we can plug this into the equation and obtain one equation one unknown. 2.74 = (X/2.22X+2650) + (X/340.3), solving for X I got 708.54m and 1570.86m. Which are way off to the 500 yards you claim. Give me your thoughts on this. 


#5
Jun2110, 08:33 PM

P: 8




#6
Jun2110, 08:38 PM

P: 59

By the way a far more accurate way to solve this problem is to simply measure muzzle exit velocity and angle of exit.
Vx=Vo*cos(theta) X=Vx*t Next time you go to the shooting range try this!!! 


#7
Jun2110, 09:39 PM

P: 66

first, you have to convert the units. you've stated you want meters/second. you forgot to convert the velocity
also, your calculations show that the distance is farther than he claimed, which is strange because he would want to exaggerate it, not downplay it. i myself did not think about graphing the points and approximating so good job on thinking of that method (if you graph it does actually look like a line) but, this equation t = X/Vb is flawed because that only works for constant velocity for situations that have changing velocity you must use the calculus definition Vb = dx/dt = 2.22X+2650 (which is wrong, units) one way is to solve this differential equation for x which involves the natural logarithm there is probably some other ways but i can't think of them right now :\ wwshr for your second post that is how you would find the initial velocity but you cannot use X=V_x*t because velocity is not constant, also within a timespan of 23 seconds the velocity changes by 2030 meters/second which is not a significant part of the initial velocity which is around 800 m/s, the bulk of which is in the x direction, so considering the x and y components seems not too necessary in this case 


#8
Jun2110, 09:53 PM

P: 59

Yes, you are right a assumed constant velocity, I cannot think of any other way of solving this without further information. Ofcourse drag is clearly playing a big part in this problem. Perhaps, dv/dt=v(dv/dx) could be useful but then again we would have to know the mathematical expression for the drag force, which could be something like 0.001V^2 or similar.
quantum13, you are right I made the mistake of not converting the units of velocity. With the new correction the answer becomes closer to 500 yards. I will not post my exact answer since I have been told is against forum rules. 


#9
Jun2110, 10:09 PM

P: 66

interesting, i looked at the equation dv/dt=v(dv/dx)
if you use that then a = 2.22v but you know a = F/m = .001v^2 so you get .001v^2 = 2.22v .001v = 2.22 v = 2220 what is this supposed to signify? 


#10
Jun2210, 08:38 AM

P: 59

I was thinking of using it to related distance to velocity, but then I realized we do not have final and initial velocities. But I worked out the equations anyways in case it is useful for future reference. In the attachment the work is shown, where Vf is final velocity and Vi is initial. X is distance traveled and m is the mass of the bullet.



#11
Jun2210, 02:15 PM

P: 66

i am so bad with pf latex... how did you make your thumbnail?
i worked out an equation for x = f(t) by integrating twice from a = dv/dt = kv^2 v = g(t) dx/dt = g(t) x = f(t) and using your equation to find a value for the constants and found that for a time of 2.74 seconds the distance was around 650 yards (i used my graphing calculator's solver program to solve, the equation was in the form x = a ln (b  cx) ,a,b,c>0 now if you take into account the fact that the muzzle was tilted by some amount, i think a distance of 500 yards is reasonable 


#12
Jun2210, 03:29 PM

P: 59

I used Microsoft Word, Microsoft Equation which is very easy to use. Is there a way to write formulas directly in the forum?
quantum13, that is a quite reasonable result. I still think with the information given, no exact answer can be obtained. 


#13
Jun2210, 04:38 PM

P: 66

sure, you can find a guide here:
http://www.physicsforums.com/showthread.php?t=8997 it's kind of hard to use for a beginner though in my opinion because if you try to "preview post" then it will sometimes display some weird thing that you already deleted as part of the code. i don't know why it does that. i think i will try microsoft's editor next time 


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