Artillery and Elevation ```Name: Jim H. Status: educator Age: 40s Location: N/A Country: N/A Date: Thursday, August 22, 2002 ``` Question: During the civil war artillery was often placed on higher points of elevation. If a field piece like an ordinance rifle at 15 degrees elevation could fire a projectile 4500 feet on "level ground", what would the range be if it were on a mountain 1500 feet above a valley? Time of flight about 5 seconds. Is there a general rule relative to increased elevation and distance gained? Replies: Jim - To solve your problem you need to convert the speed of your projectile into two vectors - one horizontal and one vertical. (a little trig would help) Lets make it easy... your fire your cannon ball in the horizontal from height of 32 feet. (a falling object accelerates at 32 feet per second per second) It would have a vertical velocity of zero when it leaves the mouth of the cannon. At the end of two seconds its velocity would be 64 feet per second - an average of 32 feet per second ( 0 + 64 divided by 2). It would hit the ground in two seconds. If you know the horizontal velocity - let us make it 1000 feet per second - you can calculate the distance traveled before striking the ground after the two seconds calculated above.... 2000 feet. The farther you tip the cannon up from vertical, the greater the vertical velocity at the mouth of the cannon (and the less the horizontal). If we could keep the 1000 feet per second horizontal velocity, and tip the cannon up so is would have a vertical velocity of 32 feet per second... it would take one second before the vertical velocity would be zero. It would then begin its decent. If it were the same as our first example, it would fly for four seconds... one second up, one second down to the original height of the cannon mouth, and the two seconds we discussed earlier. In four seconds it would travel 4000 feet horizontally. Not a totally realistic answer, because as the vertical velocity of the projectile increases the horizontal velocity will decrease... unless the total velocity (the vector sum of the H and V) increases with the tilt of the barrel. The realism of this example is also stretched because we are ignoring air resistance and the motion of air currents. It would work rather well on the moon if we used the acceleration due to gravity of about 5 feet per second per second. Larry Krengel Jim, The "rule" is that horizontal time of flight equals vertical time of flight. The time required to go up and come down, ending up 1500 feet below launch, equals the time spent moving horizontally. Vertical motion has constant downward acceleration: y=y0+{v0y)t-(1/2)gt^2. Discover the time in flight based on initial vertical velocity. Horizontal motion has constant velocity: x=x0+(v0x)t. V0x and v0y are the x and y components of the initial velocity. Analyze the level ground situation to discover what initial velocity is. It will be the same when launched from increased elevation. Dr. Ken Mellendorf Physics Instructor Illinois Central College If the shell leaves the gun moving at an angle of 15 degrees, it will arrive back at the original height also moving at an angle of 15 degrees. If the gun's height is not too much greater than the target's height, the rest of the shell's trajectory will be a small section of a parabola which you can approximate as a line inclined at 15 degrees. Tim Mooney Click here to return to the Physics Archives

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