Do the two balls meet at the halfway point?

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The discussion revolves around the motion of two balls: one dropped from the top of a building and the other thrown upwards from the ground with the same initial speed. It is established that the dropped ball starts with a speed of zero and accelerates downward, while the thrown ball starts fast and decelerates. Participants conclude that the balls will cross paths above the halfway point of the building due to their differing velocities and acceleration. To find the exact crossing point, it is suggested to use kinematic equations, treating the building height as a variable. The conversation emphasizes the importance of calculations to confirm the intuitive understanding of their meeting point.
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A ball is dropped from rest from the top of a building and strikes the ground with a speed Vf. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is Vo=Vf, the same speed with which the first ball will eventually strike the ground. Ignoring air resistance, decide whether the balls cross paths at half the height of the building, above the halfway point, or below the halfway point. Give your reasoning.

Hmm...
I thought of using kinematics to find if both times would be equal to each other when they arrive to halfway x = vt +1/2at^2, but the only difference between the balls would be their direction considering the velocities are similar.
What to do? Ty
 
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Hi LadiesMan,

LadiesMan said:
A ball is dropped from rest from the top of a building and strikes the ground with a speed Vf. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is Vo=Vf, the same speed with which the first ball will eventually strike the ground. Ignoring air resistance, decide whether the balls cross paths at half the height of the building, above the halfway point, or below the halfway point. Give your reasoning.

Hmm...
I thought of using kinematics to find if both times would be equal to each other when they arrive to halfway x = vt +1/2at^2, but the only difference between the balls would be their direction considering the velocities are similar.
What to do? Ty

You can definitely use kinematics to find the answer.

But conceptually, think about the fact that one starts out moving slow and speed up, while the other starts out moving fast and slow down. How does that affect the spot where they pass each other?
 


Thanks!

So:

Throwing up: Starts fast and slows down
Dropping: Starts slow and speeds up
In this sense, greater velocity equals greater distance traveled. Therefore the balls should cross each other above the halfway point.

Is that true?
 


LadiesMan said:
Thanks!

So:

Throwing up: Starts fast and slows down
Dropping: Starts slow and speeds up
In this sense, greater velocity equals greater distance traveled. Therefore the balls should cross each other above the halfway point.

Is that true?

That sounds right to me.
 


however, intuition often plays tricks on us, so, just for practice, you should consider doing the maths behind that and find out where exactly they meet, now since you know the fundamental difference between both motions.
 


Thanks!

@above: We are provided with only one value such that a = 9.8m/s^2. Thus, how can we calculate it's location?
 


LadiesMan said:
Thanks!

@above: We are provided with only one value such that a = 9.8m/s^2. Thus, how can we calculate it's location?

You can calculate at what fraction of the height of the building they cross (the halfway mark, 1/10 of the way up, etc.).
 


to think about it, I'm not sure how to prove that the balls would arrive at above the halfway point. any suggestions? possibly using equations?
 


LadiesMan said:
to think about it, I'm not sure how to prove that the balls would arrive at above the halfway point. any suggestions? possibly using equations?


To use equations, just set the height building to be h (or choose a numerical height) and solve the two motions for the time in which they are at the same position (calculating some needed quantities like the final speed of the dropped ball along the way). Once you find that position, you can compare it to the building height to answer the question.
 
  • #10


Hmm, but we don't know the height of the building? :S
 
  • #11


LadiesMan said:
Hmm, but we don't know the height of the building? :S

What I meant was to set the building height to be an unknown variable h, and then solve for the height at which the balls cross.

Then, just as an example, if you found the balls cross at the vertical position y=\frac{1}{4}h then you know that they cross below the midpoint of the building.
 
  • #12


Ight! Thank you! :)
 
  • #13


Glad to help!
 
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