Question

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?

Answer 1

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

$s=ut+0.5at^2$

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

Substituting

$h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9$

Distance traveled by second bullet

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

Substituting

$h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2$

Solving both equations

$600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s$

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

Height at t = 7.28 s

$h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m$

Passing of B occurs at 4108.31 height.