**Problem 1A-7*:** Two Identical Guns Along a Vertical Line

Two identical guns are mounted along a vertical line pointing toward each other.
If the guns are fired simultaneously, two bullets will collide somewhere and
smash each other into harmless dust. The position of the bullet of the first
gun that is mounted at the higher end of the vertical line is denoted as
y_{1}(t), and the position of the bullet of the second gun is denoted as
y_{2}(t) with time t set to zero when the guns are fired.
y_{1}(t) and y_{2}(t) are given as

y_{1}(t) = H - At - Bt^{2}and y_{2}(t) = At - Bt^{2}, where A, B and H are positive constants. The constant B is usually expressed as g/2 with the meaning of g becoming clear later. Here we just use the name B for simplicity of writing

(a) What are the positions and speeds of the two bullets, respectively, at the time when the
guns are fired

simultaneously?

(b) At what time do the two bullets collide?

(c) What are the speeds of the two bullets when they collide?

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**Solution:**

(a) The speeds of two bullets are

At t = 0, y_{1}(0)=H, v_{1}(0)=-A, y_{2}(0) = 0, v_{2}(0) = A. Thus, at the time when two guns are fired simultaneously, bullet 1 is at height H with a vertically downward speed of A, and bullet 2 is at the height of 0 with a vertically upward speed of A.

(b) When two bullets collide at time t, y_{1}(t) = y_{2}(t);
solving this equation, we will get the time of collision. This leads to

Thus, the two bullets collide at t = H/2A.

(c) At t = H/2A,

Thus, at the time of collision, bullet 1 is moving vertically downward with a speed of (A^{2}+BH)/A, and bullet 2 is moving vertically upward with a speed of (A^{2}-BH)/A.

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