Question

A 20 gram paintball exits a rifle at a speed of 985 m/s. The barrel is 0.8 m long. What was the force during the bullet’s exit launch?

Answer 1

Answer:

Data:)

initial velocity = 0, beacause gun is at rest

final velocity. = 300m/s

Acceleration = ?

distance = 0.9 metres

Solution:

Formula

2as= vf^2 - vi^2

2a(0.9) = (300)^2 - (0)^2

1.8a = 90000

a = 90000÷1.8

{a = 50000} Ans

Or

{5×10^4}

Explanation:

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