Freefall-Type Question (1 Viewer)

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

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.
 

Chi Meson

Science Advisor
Homework Helper
1,699
10
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."
 

HallsofIvy

Science Advisor
41,626
821
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...
 

Attachments

Nevermind, problem solved :)
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top