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 = 113 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.

Vertical Displacement Data ------> 6.25, 7.12, 7.25, 6.16 7.16, 6.07

What is the most accurate muzzle velocity that the soldier can report to his sergeant?

What is the uncertainty in this measurement?

Explanation / Answer

A.

Average Bullet drop

havg = (6.25 + 7.12 + 7.25 + 6.16 + 7.16 + 6.07)/6

havg = 6.6683 cm = 0.066683 m

using equation

y = y0 + u*t + 0.5*g*t^2

y0 = 0

u = 0

y = 0.5*g*t^2

t = sqrt (2y/g)

t = sqrt (2*0.066683/9.81)

t = 0.116597 sec

d = v*t

v =d/t = 113/0.116597 = 969.15 m/sec (Most accurate)

B.

max. velocity possible

t = sqrt (2*0.0607/9.81) = 0.11124 sec

Vmax = d/t = 113/0.11124 = 1015.82 m/sec

uncertainty in max velocity will be

dVmax = 1015.82 - 969.15 = 46.67 m/sec

min velocity possible

t = sqrt (2*0.0725/9.81) = 0.12157 sec

Vmin = d/t = 113/0.12157 = 929.51 m/sec

uncertainty in min velocity will be

dVmin = 969.15 - 929.51 = 39.64 m/sec

So total possible uncertainty will be

dV = (46.67 + 39.64)/2 = 43.155 m/sec

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