Freefall-Type Question: Solving for Ball Catching and Building Length

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Homework Help Overview

The problem involves two balls: one dropped from a building and another thrown downwards after a delay. The goal is to determine if the second ball catches the first and, if so, the minimum height of the building required for this to occur.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting equations for displacement equal to each other to find a solution. There is mention of considering initial velocities and the effect of time delay on the second ball's motion.

Discussion Status

Some participants have provided hints regarding the setup of equations and the importance of time reference points. There is acknowledgment of potential issues with calculations, particularly concerning negative values and the interpretation of results. The discussion reflects a mix of interpretations regarding the problem's requirements.

Contextual Notes

Participants note confusion around the calculations leading to negative values and the ambiguity of the provided answer indicating a height of 180m without clarity on its context. There is also a reference to imposed homework constraints that may limit the information available for solving the problem.

Iceman2032
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I am really struggling with this question:

A ball is dropped from the top of a tall building. Exactly 2 seconds later, another ball is thrown downwards from the top of the same building at 25 m/s. Will the second ball catch the first ball? If so, what is the minimum length of the building? If not, what is their closest approach?

I do not need the full solution, just a hint. It seems that whatever I do gets to a dead end. From the 5 variables (d, t, initial v, final v, and acceleration) I only have 2 of them in both cases...initial velocity and acceleration (which is 10 m/ss...given) I have also sketched the graphs but I just don't get this question.

Any help is appericiated.
 
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IF the 2nd one catches the first, then they would have the same displacement, wouldn't they? set two equations for d equal to each other and solve for time. If you get a negative answer, then "it can't happen."
 
distance fallen in t seconds is x(t)= (g/2)t2+ v0t
where v0 is the initial velocity.
The "dropped" ball has initial velocity 0, the thrown ball has initial velocity 25 (I'm taking positive downward). Also, be careful about when t= 0 is. If you take t= 0 to be when the first ball is dropped, the time the second ball is falling is t-2. The second ball will catch up to the two when they both have the same "x" so set up the two formulas, set them equal, and solve for t.
 
Okay, I took your suggestions and came up with the calculations attached to this post.

First of all excuse the typos and all, I just noticed an 'is' needs to be an 'if'. Took me some time to convert my scribbles into a neat image.

Now as you can see I have ended up with a negactive value for x. Something tells me I'm doing something really wrong. I mean the value of x is supiciously small...and negative. According to the first reply, they should never catch.

Also in my answer sheet, the answer to this problem is indicated as: 180m

It doesn't say wether 180 is their closest approach or height of the building.

So can you please tell me what I am doing wrong? Am I even using the right formula?! And after I calculate x, where do I go from there?

I really get lost in these 'catch' problems...
 

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Nevermind, problem solved :)
 

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