Falling Objects Puzzle: Find Ratio of Heights

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Discussion Overview

The discussion revolves around a physics problem involving two stones thrown horizontally from different heights, with the goal of determining the ratio of the heights of the buildings based on the horizontal distances traveled by the stones. The conversation touches on concepts of projectile motion, time of flight, and the effects of gravity, while exploring various interpretations and calculations related to the problem.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant states that since the second stone travels twice the distance, it must have fallen for twice the time, leading to a proposed height ratio of 2:1.
  • Another participant challenges this by noting that the distance fallen is proportional to the square of the time, suggesting that the relationship is more complex than initially thought.
  • A different participant calculates that the ratio of the heights is 4:1, based on the time of flight and the equations of motion.
  • Some participants express uncertainty about the assumptions made, particularly regarding the simultaneous landing of the stones and the implications of measuring time accurately.
  • One participant introduces the idea of general relativity affecting the problem, although this is presented in a light-hearted manner.
  • There is a discussion about the mathematical steps involved in deriving the height ratio, with some participants questioning specific calculations and clarifying misunderstandings.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the problem and the calculations involved. While some arrive at a height ratio of 4:1, others question the assumptions and the validity of the problem itself. The discussion remains unresolved regarding the implications of time measurement and the assumptions about simultaneous landing.

Contextual Notes

There are limitations in the assumptions made about the problem, particularly regarding the neglect of air resistance and the implications of time measurement accuracy. The relationship between time and distance in free fall is also a point of contention.

Who May Find This Useful

This discussion may be useful for students and enthusiasts of physics, particularly those interested in projectile motion, mathematical modeling, and the complexities of interpreting physics problems.

JohnnyB21
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two dimensions

Two stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands twice as far from the based of the building from which it was thrown as does the other stone. Find the ratiio of the height of the taller building to the height of the shorter building.
 
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Tip:
Horisontal velocity is constant.
distance = velocity * time
As the second stone travel twice the distance of the first one, it must bee falling for twice the time:
d=v*t(1)
2*d=v*t(2)

Hope that helped.
Cheers.
J.
 
More

If it falls twice the distance and it takes twice the time, then that means it's twice as high, right? The ratio would be 2 to 1? That seems logical but it seems too simple to be correct. What am I not seeing?
 
JohnnyB21 said:
If it falls twice the distance and it takes twice the time, then that means it's twice as high, right? The ratio would be 2 to 1? That seems logical but it seems too simple to be correct. What am I not seeing?
No one said it fell (vertically) twice the distance! It fell for twice the time. Now you have to know how distance fallen relates to time. The distance fallen is proportional to the square of the time: [itex]d = 1/2gt^2[/itex]

So... if the time to fall is double, what can you say about the relative height of the buildings?
 
this question looks tricky but if you know how to tackle it its really quite easy. Doc Al said it right. If you do the calculation, you would get this:
Time of the taller building = 2 times the time of the smaller building, since horizontal distance is double
Horizontal Distance traveled by the projectile of the taller building = 2 times smaller

Now we can work with the verticla component and time in order to get the height:
Shorter Building: d = volt + 1/2at^2
d = 1/2at^2
t = sqrt(2d/a)

Taller Building: d = volt + 1/2at^2
d = 1/2a(2t)^2
t = sqrt(8d/a)
Therefore, the ratio of building heights is 4:1.
 
Most questions like these end up with squares.
I use to think that since the vertical distance is a quadratic function of time, "2 times the time" must mean 2^2 times the distance. If the stone from the taller building flew three times the distance of the other stone, height ratio would be 3^2=9 to 1.

Assuming that we are neglecting atmospheric drag... =)

Cheers.
 
Last edited:
Isnt this a bogus question to start with? Seeing we can't measure time properly. Could it even possible for those two stones to land at a precise moment in time simultaneously?

What would the odds be on that? Can someone give me odds of two objects simultaneously touching the same point since we don't have a proper measurement of time? Would you have to go down to the quantum level to even start to guestimate?

Just thinking here, nm if you don't want to continue.
 
soln

4:1

Using parametric equations gives a simple solution

y1(t) = h - at^2/2
y2(t) = H - at^2/2

solve yn(t) = 0

t1 = sqrt(2h/a)
t2 = sqrt(2H/a)

x1(t) = vo*t
x2(t) = vo*t

r1 = vo*t1
r2 = vo*t2

r2/r1 = (vo*t2)/(vo*t1)

r2/r1 = t2/t1 and r2 = 2r1

2 = sqrt(2H/a)/sqrt(2h/a)
4 = (2H/a)*(a/2h)

4 = H/h
 
mapper said:
Isnt this a bogus question to start with? Seeing we can't measure time properly. Could it even possible for those two stones to land at a precise moment in time simultaneously?
Why do you assume that the stones land simultaneously? All we know is that the stones are thrown with the same speed. No one said they are thrown or land at the same time. And it's not relevant anyway.
 
  • #10
The stones are thrown at the same horizontal speed, but the vertical heights of the buildings are different. The time it takes for the stone to hit the floor has nothing to do with horizontal speed (now we are neglecting air resistance here), but only to do with the height with which it was dropped. Remember, gravity works in the vertical direction and thus only affects the vertical.
 
  • #11
Forgive me if this is a stupid question, but I just don't see how Nenad got:
d = 1/2a(2t)^2
t = sqrt(8d/a)

How do you get the "8" in this answer?

I am getting sqrt(d/2a)=t

I am sure there is a simple answer to this. Can anyone point it out to me?
 
  • #12
paul11273 said:
Forgive me if this is a stupid question, but I just don't see how Nenad got:
Beats me what Nenad was up to. The problem is straightforward:
[tex]D_1 = 1/2 a t_1^2[/tex]

[tex]D_2 = 1/2 a t_2^2[/tex]

So, since [tex]t_2 = 2t_1[/tex]:
[tex]D_2 = 1/2 a (2t_1)^2 = 4(1/2 a t_1^2) = 4D_1[/tex]
 
  • #13
Thanks Doc Al. That is how I worked it out and came to the 4:1 answer.
I thought I was missing something with that 8 in Nenad's intermediate steps.
 
  • #14
Now, can you figure out the answer with the corrections necessary to include the general relativity component induced by the differing heights of the buildings? <is evil>
 
  • #15
Mad Arab said:
Now, can you figure out the answer with the corrections necessary to include the general relativity component induced by the differing heights of the buildings? <is evil>

Yes, considering GR but not air resistance it's D2 = 4.000 D1 for even the tallest building (and half tallest-building) on Earth. :wink:
 

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