A soldier with a machine gun,falling from an | vedclass

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(C) To resist the downward acceleration,the upward force exerted by the bullets (thrust force) must balance the weight of the soldier and the gun.Let

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$100$

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(C) To resist the downward acceleration,the upward force exerted by the bullets (thrust force) must balance the weight of the soldier and the gun.Let $M$ be the total mass of the soldier and the gun,and $m$ be the mass of one bullet.The force exerted by the bullets is equal to the rate of change of momentum of the bullets: $F = \frac{N}{\Delta t} 	imes m 	imes v$,where $\frac{N}{\Delta t}$ is the number of bullets fired per second.Equating this force to the weight of the soldier: $Mg = \frac{N}{\Delta t} 	imes m 	imes v$.Given: $\frac{N}{\Delta t} = 40 \,s^{-1}$,$m = 49 \,g = 0.049 \,kg$,$v = 500 \,m/s$,and $g = 9.8 \,m/s^2$.Substituting the values: $M 	imes 9.8 = 40 	imes 0.049 	imes 500$.$M 	imes 9.8 = 40 	imes 24.5 = 980$.$M = \frac{980}{9.8} = 100 \,kg$.

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