A sniper shoots a bullet horizontally over level ground with a velocity of 1.22 x 10^3m/s. At the instant the bullet leaves the barrel, ?
its empty shell casing falls vertically with a velocity of 5.5m/s.
a) Neglecting air friction, how far does the bullet travel?
b) what is the vertical component of the bullet's velocity at the instant before it hits the ground?
shell casing falls vertically and strikes the ground* with a velocity of 5.5m/s
- NCSLv 71 month agoFavourite answer
I think you left a phrase out of the question -- the end should read "falls vertically AND STRIKES THE GROUND with a velocity of 5.5 m/s."
time to fall t = v / g = 5.5m/s / 9.8m/s² = 0.56 s
a) d = V*t = 1220m/s * 0.56s = 685 m
b) 5.5 m/s
same as the shell casing!
Hope this helps!
- PhilomelLv 71 month ago
How high above the ground was the muzzle of the gun when it was fired?
- oubaasLv 71 month ago
Pls post it again with the missing data since this question is not serious ....
- billrussell42Lv 71 month ago
first: how long does it take the casing to fall to the ground. But we are missing the distance above the ground the rifle is...
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- RobinLv 71 month ago
if you neglect air friction, which you cant as the bullet would go into outer space