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 = 107 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 several trials, the soldier finds that the bullet strikes the target an average of 7.7 cm, below the bullseye.

Question

jahmiya · Answer

whats the question

anonymous · Answer

What muzzle velocity should the soldier report to his sergeant?

jahmiya · Answer

Table: Vertical displacement data 
Bullet drop, h, in cm 
5.92 
7.34 
7.14 
6.16 
7.49 
6.18 
------ Find the avarage drop. 
40.23 /6 = average 6.705 cm = 0.06705 m 
Y=Yo + Voy(t) + (1/2)g(t)²===> Yo=0 Voy=0 
Y = 1/2gt² ==> Y = 0.06705 
0.06705 = 1/2 (9.8) t² 
t = 0.11697===> time of drop 
d=rt d=111m, t= 0.11697 s r=? 
r= 111/0.11697 = 948.9 m/s===> Muzzle Velocity 
Uncertainty 
at 5.92 cm velocity is 1009.85 m/s 
at 7. 49 cm veocity is 897.8 m/s 
1009.85-948.9 = 60.95 
948.9- 897.8 = 51.1 
(51.1+60.95)/2 = +/-56 m/s

anonymous · Answer

I don't understand the first half.

jahmiya · Answer

i dont understand it either my brother put it he just left ill call him and ask

anonymous · Answer

Alright thank you

rvc · Answer

draw the question's outline...

rvc · Answer

separate it into horizontal n vertical component

anonymous · Answer

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