how do I determine an arrows muzzle velocity?
96240
Use kinematics! Assuming air resistance to be negligible, as it should be for an arrow, the horizontal velocity of a projectile (i.e. an arrow) is constant. Thus, the velocity at which the arrow leaves the bow is constant after it has left the bow. Time the amount of time it takes for the arrow to reach the target after it left the bow. Then, use the fact that v=x/t to find the horizontal velocity of the arrow. Since if the arrow is fired horizontally from the bow, the only velocity imparted to the arrow from the bow is horizontal, the muzzle velocity should be equal to the horizontal velocity that the arrow has in flight. The vertical velocity of the arrow will be constantly changing through out its flight (it will have a constant downwards accelleration due to gravity of 9.81 m/s^2) but, assuming once more that the arrow is horizontal in the bow, there will be no vertical motion in the bow. I suggest using a video camera to time the amount of time it takes for the bow to reach the target after it has left the bow. That way you can use a frame by frame method to get a really precise time of flight. The distance the arrow travels during that time is really easy - it's just the distance between your bow and the target.
644478
1)Start by modelling the speed of a ball in one direction; take into account friction force (negative drag) that is equivalent to negative acceleration. Loop until speed is equal or less than to zero. Then, stop.\n\n2) Do Exactly the same, now in a 2-D space: model how'd the would move given an initial vector speed.\n\n3) Model a "Hit" on a ball: given an instant (vector) speed plus a "Hit Vector", compute the resultant vector; then use that vector as the initial speed again.\n\n4) develop a "Hit?" function that returns true if the difference between the centers of the balls is less than the diameter, or if the center of the ball is closer than a radius to a wall (the stick can be thought of as another ball)\n5) Every time Hit? is true, recompute vector speed for every involved ball.\nVoila! you have your game. Let me know if it's useful and the results.
56830
Muzzle velocity will be effected by barrel length.\nIf I remember correctly Guns And Ammo did a test a few years ago concerning this. I believe the test rifle was chambered in .308  With the loss of 2 inches of barrel length the fps drop was approx 70 fps.  You would have to go to their website and reference the article for the exact specs..
730340
Projectile range would be the equivalent of horizontal distance.\nThe forces involved are gravity (acting on the projectile's mass in the vertical direction), and the "push" from the initial thrust.\n\nAs the angle of the firing position increases above the horizontal, more of the initial push is used to combat gravity.\n\nIf the angle is too low, the projectile will meet the ground before it's maximum range (not enough action against gravity).\n\nFor most projectiles, the optimum angle is between 42 and 45 degrees (depending on the strength of the initial impetus). This is because the trajectory approximates a parabola.\n\nWith the given choices, the answer will have to be 45 degrees (answer B)...
1140159
Not sure.  I shoot a couple of long bows that were hand made by my grandfather.  Also have a re-curve, but no compounds.\nSorry, this hardly answers your question.\nGood luck.
987748
1.Compute initial velocity V0\nUsing equation S=V0*t-(g*t^2)/2 we find that the initial velocity V0=35/3 + 3g/2\n2.What ever goes up must come down V0*t=-(g*t^2)/2 t=2*V0/g\n3.Total time travel = 2t \nYou to the math\n\nBest\n\nHint:  the arrow will stop midair in about 5.3 seconds
51395
I don't know about using a pendulum, and the ballistic gelatin is too inconsistent for accurate velocity measurements.\n\nI use a chronometer. VERY simple device - there are two photoreceptors placed a set distance apart, both with white screens above them to serve as a backdrop. Any object passing across the first photoreceptor starts a timer, and the timer stops when that object passes the second receptor. Because the distance is known, the chronometer is able to calculate velocity.\n\nAn inexpensive ballistic chronometer will set you back about $90-$100. add features like multiple shot memory, printers, or large LCD displays and the sky's the limit!
754389
Learn to answer these questions yourself!\n\nIf the air is assumed to be frictionless, the vertical and horizontal velocities are independent.  The altitude is given by H = H0 + V0t + 1/2 a t^2.  H0 is +30 meters, V0 is +24.5 m/sec, and a = -9.8 m/sec^2.  If the target is at ground level, you are solving a quadratic for H = 0.\n\nSolving the quadratic gives you the time of flight, t.  Multiply the time of flight by the horizontal velocity (49 m/sec * cos(30°)) and  you get the distance of the target.\n\nOf course, you can just look at the problem and the answers and see if there's just one answer that's anywhere close.  In this case, the initial vertical speed is 24.5 m/sec; it takes 2.5 seconds for the arrow to rise to its peak altitude, and another 2.5 seconds to come back down to the altitude of the bow (5 seconds total).  Falling the final 30 meters would take on the order of a second.  The cosine of 30 degrees is about 0.87, so the horizontal velocity is about 42.5 m/sec; multiply by 6 seconds, and you get about 250 meters.  Ergo, (4) is the only answer that's even close.
622176
would depend on the mass, and air resistance of the arrow. as arrows have most of their weight at one end I guess they will fall very weell, and with quite low air resistance. a guess would be well in excess of 200 mph, a diving bird of prey can get close to this and an arrow wil have a much smaller profile, but as speed increses the gai you get will diminish. you could try putting numbers in a simulator online if you have some.(like this one below)
1114215
17 HMR (Hornady Magnum Rimfire) ammo has a 17 gr. hollow point (filled with polycarbonate) boat tail bullet with a 0.172" diameter and delivers a muzzle velocity of about 2550 fps (depending on the barrel length). There is 5.4 gr of what looks like Lil' Gun powder. The muzzle energy is 245 ft-lb and with a 1-10" twist, the bullet is spinning at 183,500 rpm. A typical 22 Long Rifle cartridge has a bullet weight of 40 gr. or more than twice the mass of the 17 HMR.
869691
a)F = MV=  m1v1 +/-  m2v2  \nusing these we get,  95*4.1 + 85*5.5 (since same direction)=857kgm/s = (95+85)* V. so, V = 4.76m/s\nb) MV = M (6) = (9500+m)*6 = 9500*16 + m*0. \nso, (9500+m) = 9500*16/6= 25333.33 kg\nso, m = 25333.33-9500 =15833.33kg\nc)work done against friction= kinetic energy of bullet\nW = FScos(tetha)  since its horizontal, angle is zero, \nand cos0 = 1, so W = FS\nwe know, coeff of friction = tangent of angle between, the frictional force  and normal reaction...\nso, the \nF = normal reaction=(1.1 +0.015)* 9.81=10.938N\nfrictional force= 0.25 *10.938 = 2.7345N\ndistance is 9.5m...\nso W = 2.7345*9.5 = 25.97 J\nthe kinetic energy of bullet has been transferred to the block...\nso the kinetic energy of bullet is 25.97J\n1/2*(0.015)* v^2 = 25.97\nso, v^2 = 25.97*2/0.015=3463.74\nso, v = 58.85m/s\nhope this helps...:)
623687
Ask at a pro shop.Find a draw weight your comfortable with. If it's a compound bow you are after. It's not just the draw weight {that can to some extent be adjusted} But also the length of your draw. Take a tape measure in your bow hand slide the other end to your anchor point. Just under your jaw beneath you ear. That will roughly measure your draw lengh. This is also adjustable on some compound bows. If the bow is the wrong weight & especialy draw lengh you'le never be able to shoot it accurately. Get some proper guidance before commiting yourself to investing in the bow. the wrong bow could not only prove useless but if second hand & faulty could be lethal, to you as well as anyone else. Kev
1130695
The muzzel velocity varies, depending on the brand of ammo, length of barrel, etc.\n\nYou could use a typical muzzel velocity of 1200 feet per second though...thats 818 mph.  I've seen manufacturers claim up to 1400 feet per second...954mph.\n\nhope this helps.
8368
Assuming Federal's published muzzle velocity of 2600 FPS from a 24" barrel, and using Sierra's published ballistic coefficient of .477, you can expect a total drop of 483.7 inches, or a little over 40 feet. Notice that the 168 goes subsonic somewhere around 900 yards.\n\n168 Grain:\nRange (yds)Vel. (fps)Total Drop\n0 2600 0\n100 2421 -2.7\n200 2249 -11.4\n300 2085 -26.9\n400 1927 -50.1\n500 1777 -83.4\n600 1637 -127.8\n700 1505 -184.4\n800 1380 -253.4\n900 1280 -348.2\n1000 1184 -458.7\n\n175 grain:\nRange (yds)Vel. (fps)Total Drop\n0 2600 0.0\n100 2428 -2.6\n200 2262 -11.3\n300 2104 -26.7\n400 1951 -49.7\n500 1805 -82.3\n600 1669 -125.8\n700 1540 -181.5\n800 1418 -249.5\n900 1315 -340.3\n1000 1219 -448.7
944006
a) Initial velocity, u , of the bullet as it hits target=1200m/s\n    final velocity, v, as it emerges =850m/s\n    distance traversed,s = thickness of target\n                                   =30cm\n                                   =0.30m\n\nv^2-u^2=2as\n\n850^2-1200^2=2a*0.30\n722500-1440000=0.6a\n-717500=0.6a\na=-717500/.6\n  =-1,195,833m/s^2\n\nv=u+at\n\n850=1200+(-1,195,833)t\n1,195,833t=1200-850\n               t=350/1,195,833    \n                =2.93*10^-4  s\n\n              \nb) F=ma\n mass of bullet, m=25grams or 0.025kg\n\nF=0.025*1,195,833\n  =29,896N\n\nc)Initial momentum of bullet=25*1200\n                                         =30,000\nInitial momentum of target=350*0\n                                          =0\n\nFinal momentum of bullet=25*850\n                                         =21,250\nFinal momentum of target=350v\n\nBased on conservation of momentum:\n\nInitial momentum of the system=Final momentum of the system\n\nSubstitute known values:\n\n30,000+0=21,250+350v\n350v=30,000-21,250\nv=8750/350\n  =25m/s
389289
You can do a comparison of the different rifle calibers using a tool at Remington.com\n\nhttp://www.remington.com/products/ammunition/ballistics/\n\nYou can compare calibers and then bullet weights as well since that's also a factor in understanding energy of the bullet at different ranges.\n\nHave fun and be safe.
905951
First of all I assume the height of \nbasket above ground is 2+3.05=5.05\nLet v be the required velocity.\nvertical component=vsin40\n=v*.766\nfrom conservation of energy\nheight gained\n=[v*.766]^2/2g=3.05\nv^2=3.05*2*9.81/.766^2\n=101.986\nv=10.09m/sec