The gun, mount, and train car of a railway had a total mass of 1.22 x 10^6 kg. The gun fired a projectile that was 80 cm in diameter and weighed 7502 kg. In the firing, the gun has been elevated 20 degrees above the horizontal. If the railway gun at rest before firing and moved to the right at a speed of 4.68 m/s immediately after firing, what was the speed of the projectile as it left the barrel (muzzle velocity)? How far will the projectile travel if air resistance is neglected? Assume that the wheel axles are frictionless.

Answers

Answer 1

Answer: 1) According to the law of conservation of momentum ..
Horiz recoil mom of gun (M x v) = horiz. mon acquired by shell (m x Vh)

1.22^6kg x 5.0 m/s = 7502kg x Vh
Vh = 1.22^6 x 5 / 7502 .. .. Vh = 813 m/s

Barrel velocity V .. .. cos20 = Vh / V .. ..V = 813 /cos20 .. .. ►V = 865 m/s

2) Using the standard range equation .. R = u² sin2θ /g

R = 865² x sin40 / 9.80 .. .. ►R = 49077 m .. (49 km)

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Explanation:

Answer:

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Explanation:

correct on edge

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