A few days ago
need help plz?
A shot fired from a gun with a muzzle velocity of 1200ft per second is to hit a target 3000 ft away. Determine the maximum angle of elevation for the gun
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A few days ago
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we know the bullet falls at 32.2 ft/s2
we want the distance dropped over the time it is in the air to be zero.
s=s0 + v0t + 1/2at^2
0 = 0 + 1200 sin(Elevation)t – 1/2(32.2)t^2
we also know the time the bullet is in the air
t = d/r = 3000/(1200 cos(Elevation))
t = 5/(2cos(E))
substitute:
0 = 0 +1200sinE(5/(2cosE)) – 16.1(5/(2cosE))^2
0 = 3000sinE/cosE – 16.1(25/4)/cosE^2
0 = 3000sinEcosE – 16.1(25/4)
since sin(2u) = 2sin(u)cos(u)
0 = 1500 sin(2E) -16.1(25/4)
sin(2E) = 16.1(25/4)/1500 = 0.0670833333
2E = sin-1(0.0670833333)
2E = 3.846
E = 1.92 degrees
or you aim 100 feet above the target(?!?!?)
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