document.write( "Question 200557This question is from textbook Trigonometry
\n" ); document.write( ": Hi,
\n" ); document.write( "I'm not sure if I'm doing this right, here's the question in its entirety: \r
\n" ); document.write( "\n" ); document.write( "The 8-in(diameter) Howitzer on the U.S. Army's M110 can propel a projectile a distance of 18,500yd. If the angle of elevation of the barrel is 45 degrees, then what muzzle velocity (in feet per second) is required to achieve this distance?\r
\n" ); document.write( "\n" ); document.write( "I'm using the formula: Vo2 sin 2(theta) = 32(distance). I've converted the 18,500yd to 55,500 feet. I don't see a use for the 8-in diameter value. \r
\n" ); document.write( "\n" ); document.write( "step 1: Vo2 sin(90) = 32(55,5000\r
\n" ); document.write( "\n" ); document.write( "step 2: Vo2 * 1 = 1,776,000\r
\n" ); document.write( "\n" ); document.write( "step 3: Vo2 = 1,776,000\r
\n" ); document.write( "\n" ); document.write( "step 4: Vo = 1,333 feet/sec\r
\n" ); document.write( "\n" ); document.write( "I don't know if I'm using the formula correctly or not. Any help would be gratefully received.\r
\n" ); document.write( "\n" ); document.write( "Thanking you in advance, Lynn
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Algebra.Com's Answer #150873 by Alan3354(69443)\"\" \"About 
You can put this solution on YOUR website!
The 8-in(diameter) Howitzer on the U.S. Army's M110 can propel a projectile a distance of 18,500yd. If the angle of elevation of the barrel is 45 degrees, then what muzzle velocity (in feet per second) is required to achieve this distance?
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\n" ); document.write( "Max distance (at 45 degs) = v^2/g (this can be derived, if you're interested)
\n" ); document.write( "d = 55500 feet
\n" ); document.write( "v^2 = 55500*32 = 1776000
\n" ); document.write( "v = ~ 1332.66 ft/sec
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\n" ); document.write( "The shell diameter doesn't matter, nor the weight of the projectile.
\n" ); document.write( "This is assuming no air friction, as usual. Physics occurs in a frictionless vacuum.
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