Posted: 11/16/2010 10:46:56 AM EDT
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My soda can coehorn mortar can shoot a can out at roughly 80 degrees and has a 9.5 second hangtime (from boom till the can hits the ground). Knowing time and de-accelleration (gravity, 9.81m/s) and neglecting air resistance, what is the max height it's reaching? )Its been soooo long since I took physics!)
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Quoted: I imagine it'd depend on how fast it's going I don't think you'd need initial velocity... all I care about is the Ymax, and since its a ballistic trajectory, we know that at the halfway point it reaches Vy=0 at 4.25 seconds... So I guess the problem could also be seen as how far does something to stop if it takes 4.25second to stop and is de-accelerating at -9.81m/s?
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Acceleration in the Y direction = -9.81
Velocity in the Y direction = -9.81*t + sin(80 degrees)*MuzzleVel Position in the Y direction = -4.405*t^2 + sin(80)*MV*t + 0 We know that PosY = 0 @ t = 0, 9.5 0 = -4.405* (9.5^2) + sin(80)*9.5*MV Solve for MV MV = 42.49 m/s PosYMax will be where VelY is zero. Solve for t when VelY is 0 0 = -9.81t + sin(80)*42.49 t = 4.27 PosY(t=4.27) = -4.405*(4.27^2) + sin(80)42.49*4.27 = 98.36 meters Did a retarded arithmetic error. Revised answer below. Now I feel like a retard. |
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AR15 math fail!!!
The OP is only interested in the max height, so it is a one dimensional problem. Neglecting resistance simplifies the problem and we can simplify it further by knowing the projectiles velocity is zero when it reaches max height at 4.25 seconds. The simplified question is when released how far does a object fall in 4.25 seconds? The acceleration is constant, y''(t)=9.81 m/s^2 Integrate once y'(t)=9.81*t+A m/s Integrate again y(t)=9.81*t^2/2+A*t+B m Boundary conditions, at t=0 y=0 and at t=0 y'=0 Solving for A and B, A=0 and B=0 y=9.81*4.75^2/2=110.67 m Edit: I am assuming the muzzle is at the same height of the ground and I can't divide 9.5 by 2 |
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Quoted: In Mortar Gunnery, that is referred to as Max Ordinate...........very critical to know during a CALFEX......hitting an inbound fast mover with a mortar round is bad for business. The Max was also taken into consideration for counterbattery, hence why we fired at lowest elevation possible. I'm sure now it makes no difference. By the mid-90s the Firefinder was pretty damn good. I'm sure now it's lots better. |
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Quoted:
the up time is shorter than the down time, I'm sure it is moving faster at the muzzle than when it hits the ground The OP specifically stated "neglecting air resistance" which means up time = down time and the muzzle velocity = impact velocity. Of course, that may not be a good assumption, given the aerodynamics of a coke can. 
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Quoted:
Quoted:
the up time is shorter than the down time, I'm sure it is moving faster at the muzzle than when it hits the ground The OP specifically stated "neglecting air resistance" which means up time = down time and the muzzle velocity = impact velocity. Of course, that may not be a good assumption, given the aerodynamics of a coke can. considering the very low velocity and the large mass of the can, it might not be such a bad assumption note how three guys did the same bone simple freshman level calc and got three different answers if an object is accellerated from zero velocity by gravity for 4.75 seconds it will go 110.6 meters. that's yer answer |
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Quoted:
Acceleration in the Y direction = -9.81 Velocity in the Y direction = -9.81*t + sin(80 degrees)*MuzzleVel Position in the Y direction = -4.405*t^2 + sin(80)*MV*t + 0 We know that PosY = 0 @ t = 0, 9.5 0 = -4.405* (9.5^2) + sin(80)*9.5*MV Solve for MV MV = 42.49 m/s PosYMax will be where VelY is zero. Solve for t when VelY is 0 0 = -9.81t + sin(80)*42.49 t = 4.27 PosY(t=4.27) = -4.405*(4.27^2) + sin(80)42.49*4.27 = 98.36 meters Horizontal vectors don't apply in this case (we are assuming no friction). The only force affecting hang time is gravity, and we know what that equation is (see my last). Let's say you're on the moon. You shoot a bullet and at the exact same time drop a feather. Which hits the ground first? |
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Quoted:
Quoted:
Acceleration in the Y direction = -9.81 Velocity in the Y direction = -9.81*t + sin(80 degrees)*MuzzleVel Position in the Y direction = -4.405*t^2 + sin(80)*MV*t + 0 We know that PosY = 0 @ t = 0, 9.5 0 = -4.405* (9.5^2) + sin(80)*9.5*MV Solve for MV MV = 42.49 m/s PosYMax will be where VelY is zero. Solve for t when VelY is 0 0 = -9.81t + sin(80)*42.49 t = 4.27 PosY(t=4.27) = -4.405*(4.27^2) + sin(80)42.49*4.27 = 98.36 meters Horizontal vectors don't apply in this case (we are assuming no friction). The only force affecting hang time is gravity, and we know what that equation is (see my last). Let's say you're on the moon. You shoot a bullet and at the exact same time drop a feather. Which hits the ground first? He's implicitly using the vertical vectors. He's actually trying to be more thorough, because he's calculating, rather than assuming, when the time to apogee is. Looks like he just screwed up the math somehow calculating that time. |
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It will go up faster than it comes down, so what you really need to know, is how long it goes up before it comes down. NO WAY are this times the same and since the angle is 80 the can will fall faster than gravity. If it doesn't why do we shoot down out of plane, why not just drop the bullets. |
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Quoted:
Horizontal vectors don't apply in this case (we are assuming no friction). The only force affecting hang time is gravity, and we know what that equation is (see my last). Let's say you're on the moon. You shoot a bullet and at the exact same time drop a feather. Which hits the ground first? When is the last time anyone else in this thread took a physics class? He didn't shoot the cannon straight up, he said he fired it at 80 degrees. That means the vertical component of the muzzle velocity is sin(80 degrees) times the muzzle velocity. Every single person who has given the 110 meters answer so far has been assuming that he fired the mortar straight up. Fucking division, how does it work? |
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Quoted:
Quoted:
Horizontal vectors don't apply in this case (we are assuming no friction). The only force affecting hang time is gravity, and we know what that equation is (see my last). Let's say you're on the moon. You shoot a bullet and at the exact same time drop a feather. Which hits the ground first? When is the last time anyone else in this thread took a physics class? He didn't shoot the cannon straight up, he said he fired it at 80 degrees. That means the vertical component of the muzzle velocity is sin(80 degrees) time the muzzle velocity. Every single person who has given the 110 meters answer so far has been assuming that he fired the mortar straight up. In this case, all that matters is the vertical component of the muzzle velocity. A mortar shot straight up and a mortar shot at 80* that both have the same flight time will produce the same apogee height, although the 80* mortar will need a higher muzzle velocity to do so. Since we only care about the apogee height and not the muzzle velocity, either method works fine. You have apparently made an arithmetic error though, since your time to apogee is not one half of the flight time. ETA: You divided 9.81/2 incorrectly. When you do that correctly, your method yields a muzzle velocity of 47.31m/s, a time to apogee of 4.75s and a height of 110m.
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Quoted:
With an 80 degree bore axis at firing, how far downrange does the can land? I'm not taking that bait, I'll make 2 * 2 = 14 or friggin' something, this time. 
But anyway, if what I found for the muzzle velocity in my second attempt to not be a friggin' retard is correct... Vx = cos(80deg) * 47.32 = 8.22 m/s 8.22 m/s * 9.5s flight time = 78.1 meters |
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Quoted:
Quoted:
In Mortar Gunnery, that is referred to as Max Ordinate...........very critical to know during a CALFEX......hitting an inbound fast mover with a mortar round is bad for business. The Max was also taken into consideration for counterbattery, hence why we fired at lowest elevation possible. I'm sure now it makes no difference. By the mid-90s the Firefinder was pretty damn good. I'm sure now it's lots better. If you knew what we could do with them now, you would giggle. |
