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12.57 milesFirst of all this can't be right. I used an equation to find out how high biohazard can shoot a 1 oz projectile when we get it finished.
using hgdt at a 10X mix I got 2061 feet persecond. I used this equation from my math book. H=(Vo^2(sin A)^2)/64 here is my work. H=(2061^2(sin 90)^2)/64 H=(4247721*1)/64 H=66370 feet H=12.57 miles so is this right? It must not take air resistance and gravity into account. 12.57 miles is a long way up.
I don't think it's right because of 2 things. 1) It doesn't factor in air resistance and 2) a supersonic projectile will very quickly go subsonic. If it were correct, though, that'd be pretty crazy!
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well if it didnt take gravity into account it would travel forever, but then again I dont see any number that resembles the effect of gravity...just glancing at the equation though.. the only number that is close is 64 which is double that of which gravity makes things accelerate(32 ft/s^2).
Gravity accelerates objects toward the center of the Earth at 32.2 ft per second per second (which can also be written as 32 ft/s^2). In other words, an object's velocity will increase by 32.2 ft/s (or 9.8 m/s) for each second the object falls until it reaches its terminal velocity. Since the object isn't falling but going up terminal velocity isnt taken into consideration. However the force of gravity should always be taken into consideration even if the object is going straight up. Also with a 1oz object being fired at such a high velocity you will expierence a huge air friction force. I would suggest doing a force diagram to figure the exact number.
Use H=1/2(Vi+Vf)t
Where H= Height Vi= initial velocity Vf=Final Velocity, in this case 0 t= time or H=ViT + 1/2aTSquare a= Gravitational acceleration, 9.8 m.s
you really have to take air resistance into consideration though. espeically with a 1oz projectile.
Have you tried just using HGDt's external ballistic calculator (under the Tools menu). It should do a heck of a lot better job than the equation you are using.
BTW, the equation you are using must be taking gravity into account otherwise the shell would never decelerate and "H" would be infinite. Like David said, the 64 in the equation is probably twice the acceleration due to gravity. EDIT BTW2: You need to take air resistance into account not just because of the light ammo but because of the very high velocities. A supersonic round will slow down because of air drag very quickly regardless of how light it is.
If you would launch your projectile straight up, and not take air resistance into account, and calculate using energy equations you would get this:
1/2mv^2 = mgh Divide by m. 1/2v^2 = gh v= 2061 ft/s = 628 m/s g=9.81 0,5 * 628 * 628 = g*h 197192 = 9.81*h 197192/9.81= 20101 h = 20101 m 20101 meters = 12.49 miles! So yes, it is right. But air resistance is INSANE at 628 m/s and thus, it will not come that far.
I think your equation is just one of the basic constant acceleration equations used in physics solved for maximum height. The equation would be Vf^2=Vo^2+2ax, which gives your equation when you say Vf=0, a=32 ft/s^2, and you solve for x (and you assume a vertical shot and don't worry about the sin^2 part). This equation does model gravity, but, as others have said, neglects air resistance, which is quite significant. Just goes to show the hazards of oversimplified models.
Re: 12.57 miles
Firing this cannon in space are we
 
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