Velocity of bullet fired from a gun

  • Thread starter Kev1n
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  • #1
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1. Anyone interested in looking over the question / coments, thanks. An object mass 0.6kg is fired vertically upward in thebarrel of a gun by a vertical force of 50N acting over a period of 0.5 seconds.
Calculate
A. The velocity of the projectile when it leaves the barrel
B. Height reached by object
c. Time taken from the projectile leaving the gun to return to same height




2. a=f/m, V=Vo + AT, V^2=U^2+2AS



3. A. First velocity, A=F/M, 50/0.6 = 83.33m/s^2
Velocity, V=Vo+AT, V=0+83.33 x 0.5 = 41.67ms

B. Height: U=41.67, Vo=0, A=-G = -9.81m/s
0=1736.4-19.62xS
S=1736.4/19.62 = 88.5m

C. S=(T/2)(U+V), 88.5 = T/2 (41.67+0)
T=88.5 x 2 x 2/41.67 = 4.25 seconds
Time to return = 2 x 4.25 = 8.5 seconds
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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Your calculations are correct.
 

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