Kohul asked in Science & MathematicsPhysics · 3 months ago

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

5 Answers

  • NCS
    Lv 7
    3 months ago
    Favourite 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!

  • 3 months ago

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

  • oubaas
    Lv 7
    3 months ago

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

  • 3 months 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
    3 months ago

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

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