At t=0 bullet a is fired vertically with an initial (muzzle) velocity of 450 m/s. when 3s. bullet b is fired upward with a muzzle velocity of 600 m/s. determine the time after a is fired, as to when bullet b passes bullet a. at what altitude does this occur?

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

Solving both equations

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

Passing of B occurs at 4108.31 height.

it takes about 30 days for it to break down