How does halving the time affect the distance traveled by a falling object?

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Homework Help Overview

The discussion revolves around a problem involving the distance traveled by a falling object, specifically how halving the time affects the distance covered. The context is set within the framework of classical mechanics, particularly focusing on the equations of motion under uniform acceleration due to gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the distance covered by a falling object in different time intervals, questioning the assumptions made about gravity and air resistance. There are discussions about the interpretation of the problem statement and the translation of terms used.

Discussion Status

The discussion is active, with participants providing insights into the assumptions regarding gravity and air resistance. Some participants have offered clarifications about the nature of the problem and the potential misinterpretations of the original question. There is a recognition of differing opinions on the distance covered in the first half of the time interval.

Contextual Notes

Participants note that the problem may involve assumptions about the environment in which the object is falling, such as whether it is on Earth or another celestial body. There is also mention of the potential impact of air resistance on the acceleration of the object.

M. M. Fahad Joy
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Homework Statement



If a falling object overcome 72 metres in 6 second, how much distance it overcame in first 3 second.

Homework Equations


s = ut+(at^2)/2

The Attempt at a Solution


Here,
s = 72 m
u = 0
t = 6s
a = ?
We know,
s = ut+(at^2)/2
Or, 72 = 0*6 + (a*6^2)/2
Or, 72 = 36a/2
Or, 72 = 18a
Or, a = 4 m/s

In the second part,
u = 0
t = 3s
a = 4 m/s^2
s = ?
Again,
s = ut+(at^2)/2
= 0*t + (4*3^2)/2
= (4*9)/2
= 36/2
= 18 m (ans.)
 
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M M FAHAD JOY said:

Homework Statement



If a falling object overcome 72 metres in 6 second, how much distance it overcame in first 3 second.

Homework Equations


s = ut+(at^2)/2

The Attempt at a Solution


Here,
s = 72 m
u = 0
t = 6s
a = ?
We know,
s = ut+(at^2)/2
Or, 72 = 0*6 + (a*6^2)/2
Or, 72 = 36a/2
Or, 72 = 18a
Or, a = 4 m/s

In the second part,
u = 0
t = 3s
a = 4 m/s^2
s = ?
Again,
s = ut+(at^2)/2
= 0*t + (4*3^2)/2
= (4*9)/2
= 36/2
= 18 m (ans.)
You have apparently assumed the falling object is on another planet or somewhere in space above the Earth's surface where the acceleration of gravity is not the same as it is at Earth's surface. Since a object would fall more than 72m in 6s near Earth's surface, your assumption seems reasonable, and you have correctly worked out the result based on that assumption. But are your sure that was what was intended in the original question?
 
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tnich said:
You have apparently assumed the falling object is on another planet or somewhere in space above the Earth's surface where the acceleration of gravity is not the same as it is at Earth's surface. Since a object would fall more than 72m in 6s near Earth's surface, your assumption seems reasonable, and you have correctly worked out the result based on that assumption. But are your sure that was what was intended in the original question?
It's from my school exam question.
It's about falling object on Earth. But the gravity of Earth is 9.8 as I know.
We know that the gravity doesn't depend on mass. But for the friction of air reduces the acceleration of gravity. So it a problem me seeing the first law of falling bodies.
 
M M FAHAD JOY said:
It's from my school exam question.
It's about falling object on Earth. But the gravity of Earth is 9.8 as I know.
We know that the gravity doesn't depend on mass. But for the friction of air reduces the acceleration of gravity. So it a problem me seeing the first law of falling bodies.
The force that air resistance applies to a falling body (called drag) depends on the speed of the falling body. So you could not solve the problem by assuming drag is constant. You might look at the original problem statement and if there is anything you missed.
 
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tnich said:
The force that air resistance applies to a falling body (called drag) depends on the speed of the falling body. So you could not solve the problem by assuming drag is constant. You might look at the original problem statement and if there is anything you missed.
I have checked it again. It is ok.
 
M M FAHAD JOY said:
I have checked it again. It is ok.
In the problem statement you have written "overcome [a distance]". That is not a common expression. I assume that is your translation of the original problem. What do you think it means?
 
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tnich said:
In the problem statement you have written "overcome [a distance]". That is not a common expression. I assume that is your translation of the original problem. What do you think it means?
I am not so well in English. I have translated it from my own language. That's the problem.
 
M M FAHAD JOY said:
I am not so well in English. I have translated it from my own language. That's the problem.
Do you think the object could have started out going upward, reached a maximum height and then fallen?
 
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tnich said:
Do you think the object could have started out going upward, reached a maximum height and then fallen?
I didn't mean to criticize your English. I meant to ask if "overcome 72 m" could mean something different than "fall 72m".
 
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  • #10
tnich said:
I didn't mean to criticize your English. I meant to ask if "overcome 72 m" could mean something different than "fall 72m".
Yes, the translation will be fall 72 metres.
By the way, it was an MCQ.
a) 36m b) 24m c) 18m d) 8m

Most of my friends answered 36 metres. But I have answered 18 metres. That's my confusion.
 
  • #11
M M FAHAD JOY said:
Most of my friends answered 36 metres.
Then they fell into the trap.
Moving half the distance in half the time would be right for uniform velocity, but for uniform acceleration from rest it is a quadratic, so a quarter of the distance in half the time.
 
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  • #12
haruspex said:
Then they fell into the trap.
Moving half the distance in half the time would be right for uniform velocity, but for uniform acceleration from rest it is a quadratic, so a quarter of the distance in half the time.
Are you sure, quarter of the distance in half time?
 
  • #13
M M FAHAD JOY said:
Are you sure, quarter of the distance in half time?
½at2. What happens to that if you halve t? Note, this is only for falling from rest.
 
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  • #14
haruspex said:
½at2. What happens to that if you halve t? Note, this is only for falling from rest.
Ok, Thanks.
 

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