SOLUTION: Hi, I'm not sure if I'm doing this right, here's the question in its entirety: The 8-in(diameter) Howitzer on the U.S. Army's M110 can propel a projectile a distance of 18,500

Question 200557This question is from textbook Trigonometry
: Hi,
I'm not sure if I'm doing this right, here's the question in its entirety:
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?
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.
step 1: Vo2 sin(90) = 32(55,5000
step 2: Vo2 * 1 = 1,776,000
step 3: Vo2 = 1,776,000
step 4: Vo = 1,333 feet/sec
I don't know if I'm using the formula correctly or not. Any help would be gratefully received.
Thanking you in advance, Lynn This question is from textbook Trigonometry

Answer by Alan3354(69443) (Show Source): 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?
-----------------
Max distance (at 45 degs) = v^2/g (this can be derived, if you're interested)
d = 55500 feet
v^2 = 55500*32 = 1776000
v = ~ 1332.66 ft/sec
--------------
The shell diameter doesn't matter, nor the weight of the projectile.
This is assuming no air friction, as usual. Physics occurs in a frictionless vacuum.
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