Answer:
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.
Answer:
The weight acting upon a object
Explanation:
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Hi there!
Recall that:
Change in momentum = mass × change in velocity
Or:
Δp = mΔv = m(vf - vi)
Plug in the given values. We can assign east to be positive and west to be negative in this instance (Velocity is a vector with direction).
Thus:
Δp = (1)(-21 - 10) = -31 kgm/s OR 31 kgm/s WEST.
The correct answer is B.
Change in momentum is EQUIVALENT to the quantity of IMPULSE.
The correct answer is H.