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?

Update:

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

Relevance
• NCS
Lv 7
3 months ago

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...