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?

Update:

shell casing falls vertically and strikes the ground* with a velocity of 5.5m/s

5 Answers

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  • NCS
    Lv 7
    1 month ago
    Favorite 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

    and so

    a) d = V*t = 1220m/s * 0.56s = 685 m

    b) 5.5 m/s

    same as the shell casing!

    Hope this helps!

  • 1 month ago

    How high above the ground was the muzzle of the gun when it was fired?

  • oubaas
    Lv 7
    1 month ago

    Pls post it again with the missing data since this question is not serious ....

  • 1 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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  • Robin
    Lv 7
    1 month ago

    if you neglect air friction, which you cant as the bullet would go into outer space

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