A bullet leaves a rifle with a muzzle velocity of 521 m/s. while accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration)

$velocity=\frac{distance}{time}\\\\ velocity*time=distance\\\\ time=\frac{distance}{velocity}\\\\ time=\frac{0,840m}{521\frac{m}{s}}=0,0016s\\\\ acceleration=\frac{velocity}{time}=\frac{521}{0,0016}=325625\frac{m}{s^2}$

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gfgggdgd gfgggdgd · Answer 2 · 2022-05-31T12:19:34-04:00

Answer:

a = 1.62105 m /s2

Explanation:

25.28 m2/s2 = vi2

vi = 5.03 m/s

To find hang time, find the time to the peak and then double it.

vf = vi + at

0 m/s = 5.03 m/s + (-9.8 m/s2)tup

-5.03 m/s = (-9.8 m/s2)tup

(-5.03 m/s)/(-9.8 m/s2) = tup

tup = 0.513 s

hang time = 1.03 s

Return to Problem 11

Given:

vi = 0 m/s

vf = 521 m/s

d = 0.840 m

Find:

a = ??

vf2 = vi2 + 2ad

(521 m/s)2 = (0 m/s)2 + 2(a)(0.840 m)

271441 m2/s2 = (0 m/s)2 + (1.68 m)a

(271441 m2/s2)/(1.68 m) = a

a = 1.62105 m /s2