A soldier is tasked with measuring the muzzle velocity of a new rifle. Knowing t

Question

A soldier is tasked with measuring the muzzle velocity of a new rifle. Knowing the principles of projectile motion, he decides to perform a simple experiment at the indoor firing range. The soldier hangs a target a distance of d = 104 m from the end of the barrel. The rifle is mounted so that the bullet exits moving horizontally at the same height as the bullseye. After 6 trials, the soldier tabulates the values he measured for the distance, h, from the bullseye to the bullet strike bullet drop h in cm is 6.03,7,23,7.25,6.05,7.27,6.29.

Explanation / Answer

by using,

s=ut +1/2at^2

s=0+1/2at^2

t=sqrt(2s/g)

and

muzzle velocity v= d/t

here d=104 m

so,

v1=104/sqrt(2*0.0603/9.8)=937.78 m/sec

v2=104/sqrt(2*0.0723/9.8)=856.67 m/sec

v3=104/sqrt(2*0.0725/9.8)=855.05 m/sec

v4=104/sqrt(2*0.0605/9.8)=936.09 m/sec

v5=104/sqrt(2*0.0727/9.8)=853.85 m/sec

v6=104/sqrt(2*0.0629/9.8)=918.72 m/sec

therefore

Vavg=(937.78+856.67+855.05+936.09+853.85+918.72)/6

=893.0266 m/sec ...is answer

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