Mechanics problem — One body thrown vertically and one thrown horizontally

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Homework Help Overview

The discussion revolves around a mechanics problem involving two bodies: one thrown vertically and the other horizontally. Participants are tasked with determining the speed of the horizontal body, B, in relation to the vertical body, A, as they fall under gravity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conditions under which the two bodies come closest together and the implications of different initial speeds. There are attempts to apply the Pythagorean theorem to find the distance between the two bodies at the moment body B reaches the ground. Questions arise regarding the dimensional consistency of the equations presented and the interpretation of the problem's requirements.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants have offered algebraic expressions and calculations, while others question the validity of these approaches and seek clarification on the definitions and assumptions being used.

Contextual Notes

There is a noted complexity in the problem, with participants expressing uncertainty about the setup and the relationships between the variables involved. The discussion includes references to specific equations and the need for dimensional consistency, indicating a focus on the mathematical formulation of the problem.

  • #31
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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  • #32
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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  • #33
PeroK said:
If that is the textbook solution in post #9, then you and I were right about the interpretation of the problem.
And the question should have said the distance is "at its minimum just as..."
 
  • #34
archaic said:
Is this the textbook's solution? Weird, my solution doesn't depend on ##v_1## at all, rather on ##h,\,g## and ##x_B(t=0s)##.
I am obtaining a similar solution. There is constant relative velocity between the two projectiles.
 
  • #35
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.
 
  • #36
akawter said:
A at that time you calculated?
Please use the reply button so that readers know what post you are responding to.
 

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