# At bullet A is fired vertically with an initial 2

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

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#### Solution:

Let us write two separate kinematics equations for bullet A and bullet B. We will use the following equation:

$s=s_0+v_0t+\dfrac{1}{2}at^2$

(Where $s$ is final displacement, $s_0$ is initial displacement, $v_0$ is initial velocity, $t$ is time, and $a$ is constant acceleration)

Remember that our acceleration is -9.81 m/$s^2$ since that’s the acceleration due to gravity. It’s negative because we are assuming the upward direction to be positive. Substitute the values we know:

Bullet A:

$s=s_0+v_0t+\dfrac{1}{2}at^2$

$s=0+(450)(t)+\dfrac{1}{2}(-9.81)t^2$ (eq.1)

Bullet B:

$s=s_0+v_0t+\dfrac{1}{2}at^2$

$s=0+(600)(t-3)+\dfrac{1}{2}(-9.81)(t-3)^2$ (eq.2)

(Remember that bullet B is fired 3 seconds after bullet A. Since our initial time reference is with respect to bullet A, bullet B is 3 seconds later. Also note that both bullet A and B are fired from the same height, which we will consider to be our origin, thus $s_0=0$ m.)

To figure out when the bullets pass each other, we must note that at that point, the height of bullet A must equal the height of bullet B. Thus, we can equal eq.1 to eq.2:

$(450)(t)+\dfrac{1}{2}(-9.81)t^2=(600)(t-3)+\dfrac{1}{2}(-9.81)(t-3)^2$

Solving for t yields:

$t=10.28$ s

The height at which the bullets are equal can be found by substituting the value of t we found to either eq.1 or eq.2. They will both yield the same answer.

$s=0+(450)(t)+\dfrac{1}{2}(-9.81)t^2$

$s=0+(450)(10.28)+\dfrac{1}{2}(-9.81)(10.28)^2$

$s=4107.6$ m

$t=10.28$ s
$s=4107.6$ m