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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Let us write two separate kinematics equations for bullet A and bullet B. We will use the following equation:
(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:
(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:
Solving for t yields:
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.
Glad it helped 🙂
Could we have said that bullet A went before and made it t+3 for bullet A
So 600t – 4.9t^2 = 450(t+3) -4.9(t+3)^2
We’d get t as 7.575 which is 3 seconds less, if someone could please explain why we add 3 second to that to get it to 10.757
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. It’s all a matter of what your initial frame of reference is. Please see: https://www.youtube.com/watch?v=FsGBUM5o2-k thanks!