Mechanics problem — One body thrown vertically and one thrown horizontally

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Baiatul122001 said:
Maybe the textbook's solution is wrong

If that is the textbook solution in post #9, then you and I were right about the interpretation of the problem.
 
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Baiatul122001 said:
Maybe the textbook's solution is wrong

The solution you posted in post #9 looks wromg to me. The ##v_1^2## term in the square root should be ##-v_1^2##.

Note that for any ##v_1## the distance initially decreases. But, if ##v_1## is large, then the minimum distance occurs for small ##t##. If ##v_2## is also large, then B soon passes under A and, again, the local minimum has passed.

This ties in with what I posted earlier about there being a maximum value of ##v_1## for a solution.

The other problem with the solution is that if the minimum occurs after B has hit the ground, then when B hits the ground is still a minumum for the allowed motion. That is to say: the distance decreases until B hits the ground (which is then the minimum), but it wouldn't be the minimum if B was allowed to fall further.

This leads to an inequality for ##v_2##.

$$\frac{l}{2}\sqrt{\frac{g}{2h}} - (\frac{l^2g}{8h} - v_1^2)^{1/2} \le v_2 \le \frac{l}{2}\sqrt{\frac{g}{2h}} + (\frac{l^2g}{8h} - v_1^2)^{1/2} $$
Note that this also works in the degenerate case where ##v_1 = 0##.
 
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neilparker62 said:
I am obtaining a similar solution. There is constant relative velocity between the two projectiles.
What interpretation of the question are you using? Based on the book solution I believe @PeroK's must be right, and I confirm his analysis in post #32.