# Falling objects puzzle

1. Jul 28, 2004

### JohnnyB21

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.

2. Jul 29, 2004

### salamander

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.

3. Jul 29, 2004

### JohnnyB21

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?

4. Jul 29, 2004

### Staff: Mentor

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: $d = 1/2gt^2$

So... if the time to fall is double, what can you say about the relative height of the buildings?

5. Jul 29, 2004

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 travelled 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 = vot + 1/2at^2
d = 1/2at^2
t = sqrt(2d/a)

Taller Building: d = vot + 1/2at^2
d = 1/2a(2t)^2
t = sqrt(8d/a)
Therefore, the ratio of building heights is 4:1.

6. Jul 29, 2004

### salamander

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: Jul 29, 2004
7. Aug 5, 2004

### mapper

Isnt this a bogus question to start with? Seeing we cant 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 dont 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 dont want to continue.

8. Aug 7, 2004

### abercrombiems02

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

9. Aug 8, 2004

### Staff: Mentor

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. Aug 8, 2004

### thermodynamicaldude

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. Aug 11, 2004

### paul11273

Forgive me if this is a stupid question, but I just don't see how Nenad got:
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. Aug 11, 2004

### Staff: Mentor

Beats me what Nenad was up to. The problem is straightforward:
$$D_1 = 1/2 a t_1^2$$

$$D_2 = 1/2 a t_2^2$$

So, since $$t_2 = 2t_1$$:
$$D_2 = 1/2 a (2t_1)^2 = 4(1/2 a t_1^2) = 4D_1$$

13. Aug 11, 2004

### paul11273

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. Aug 13, 2004

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. Aug 13, 2004

### Gokul43201

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