Falling object - relating distance interval to time interval

In summary, an object falls from the top of a 103.6m high building, with a vertical distance of 4.0m between consecutive windows. The object reaches window B 0.1s after reaching window A. Through calculations, it is determined that window A is the 6th storey from the top.
  • #1
jemerlia
28
0

Homework Statement



An object falls from the top of a 103.6m high building. The vertical
distance between two consecutive windows A and B is 4.0m. The first window/storey is
also 4m from the top of the building. The object reaches window B 0.1s after it has reached window A.

Which storey is A ?

- Top of building, 103.6m above ground
|
| 4m
|
= Window
|
| 4m
|
= Window
|
|4m
|
= Window
.
.
= Window A
|
|4m,time to fall from A to B is 0.1s
|
= Window B
|
|
|
.
.
_ Ground


Homework Equations


d= 1/2gt2


The Attempt at a Solution


The height of the building seems irrelevant (it is not a multiple of 4) except to indicate the object will be falling quickly enough to satisfy the conditions of the problem. What appears to be required is to relate the time interval (0.1s) to the distance interval 4n and 4n+1

The time for an object to fall a given distance from a start velocity of 0m.s will be
SQRT(2D/g)

The most promising approach appeared to be to relate the distance between the windows A and B expressed as 4n metres to the time interval and solve for n (the number of storeys from the top) so that:
time interval = time to reach A - time to reach B

Letting DB = distance from top to Window B, DA=distance from top to Window A

0.1 = SQRT(2DB/g) - SQRT(2DA/g)

Substituting DB = 4(n+1) metres, DA = 4n metres

0.1 = SQRT(2(4(n+1))/g) - SQRT(2(4n)/g)

Sadly I'm in error because the n's cancel!

What have I done wrong?
 
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  • #2
0.1 = SQRT(2(4(n+1))/g) - SQRT(2(4n)/g)
Wright this equation as
0.1 + SQRT(2(4n)/g) = SQRT(2(4(n+1))/g).
Square both sides and solve for n.
From the height of building it is clear the last window from the top must be 25th.
 
  • #3
Still puzzled...
n should give the number of storeys A is from the top...

Following your suggestion, I have:
0.12+ (8n)/g = (8n+8)/g

then
0.01g + 8n = 8n +8

The n's still cancel - I've obviously done something wrong

The expected answer is 6
 
  • #4
0.1^2+ (8n)/g = (8n+8)/g
This step is wrong.
It should be
0.1^2+ (8n)/g + 2*0.1*(8n/g)^1/2= (8n+8)/g
 
  • #5
Please check my working - I don't get the specified answer - but it's possible that answer counted the storeys from the ground up...

I now see my mistake in squaring one side of the expression... starting from there:

0.1^2+8n/g+(0.2 x SQRT(8n/g)) = (8n+8)/g

Multiply both sides by g

0.01g +8n +(0.2g x SQRT(8n/g)) = 8n + 8

Cancel the 8n terms

0.01g + (0.2g x SQRT(8n/g)) = 8

Re-arrange:

0.2g x SQRT(8n/g) = 8-0.01g

SQRT(8n/g) = (8-0.01g)/0.2g

8n/g = ((8-0.01g)/0.2g)^2

n = (g x ((8-0.01g)/0.2g)^2)/8

Taking g to be 9.81ms^-2

Gives n as 19.91 (4sfs) so I guess n = 20
 
  • #6
Yes. n is correct.
As I told you before, from the height of building (103.6 m) it is clear the last window from the top must be 25th. Upto A number of windows 20. So below A there are 5 windows. Hence A must be 6th storey.
 

Related to Falling object - relating distance interval to time interval

1. What is the relationship between distance interval and time interval for a falling object?

The relationship between distance interval and time interval for a falling object is that they are directly proportional. As the time interval increases, the distance interval also increases at a constant rate.

2. How can we calculate the time interval for a falling object?

The time interval for a falling object can be calculated using the formula t = √(2h/g), where t is the time interval, h is the height of the object, and g is the acceleration due to gravity (9.8 m/s²).

3. Is the distance interval the same for all falling objects?

No, the distance interval for a falling object varies depending on factors such as the initial height, air resistance, and mass of the object. Objects with a larger mass will have a greater distance interval than objects with a smaller mass.

4. How does air resistance affect the distance interval of a falling object?

Air resistance can decrease the distance interval of a falling object. As the object falls, it experiences a force from the air pushing against it, known as air resistance. This force can slow down the object and reduce its distance interval.

5. Can the distance interval and time interval be used to determine the speed of a falling object?

Yes, the speed of a falling object can be calculated by dividing the distance interval by the time interval, using the formula v = d/t, where v is the speed, d is the distance interval, and t is the time interval. This will give the average speed of the object during its fall.

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