Uniformly accelerated motion under gravity

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SUMMARY

The discussion centers on uniformly accelerated motion under gravity, specifically addressing misconceptions about the velocity of a falling body. Participants clarify that while distances traveled in the first three seconds of free fall are in the ratio 1:3:5, the speed gained is consistently 10 m/s, not 5 m/s at any point. The average speed during the first second can be 5 m/s, but this does not apply to the last second of motion. The conversation emphasizes understanding motion as a continuous rate of change rather than discrete average speeds.

PREREQUISITES
  • Understanding of basic kinematics principles
  • Familiarity with the concept of uniform acceleration
  • Knowledge of gravitational acceleration (g = 10 m/s²)
  • Ability to interpret motion equations and graphs
NEXT STEPS
  • Study the kinematic equations for uniformly accelerated motion
  • Learn about the implications of gravitational acceleration on different objects
  • Explore the concept of instantaneous velocity versus average velocity
  • Review physics textbooks that cover motion under gravity, such as "Aakash module (Target 1)"
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Students of physics, educators teaching kinematics, and anyone interested in understanding the principles of motion under gravity.

Vivan Vatsa
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Uniformly accelerated motion under gravity:- I have been taught that when a body accelerates, it travels on ratio of 1:3:5... so on. I have also been taught that at the last second, anybody in the world will travel with the velocity of 5 m/s.
Why is that so?
Why I should believe on some axiom.
 
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Hello Vivan, :welcome:
Vivan Vatsa said:
I have been taught that when a body accelerates, it travels on ratio of 1:3:5... so on
This sounds very unbelievable to me. Can you specify where you picked this up and how exactly it was formulated ?
 
It is written in many textbooks I have referred. Every where it is explained in the same way. It tells that though respective of how much the body travelled, it will travel with a velocity of 5 m/s in the last second.
I don't understand why is that so...?
 
You did not referred any. Give at least one example, with title, author, page. This is what it means to give a reference.

The distances traveled during each of the first three seconds (for a body falling from rest) are indeed in the ratio 1:3:5. Their actual values are 5 m, 15m amd 25 m (taking g=10 m/s).

The speed gained during each second, including the last second, is 10 m/s.

But the speed "at the last second" is meaningless so it does not even make sense to argue if it's 5m/s or not.During the last second, as during any other second, the speed increases, by 10 m/s. So at the beginning of the last second the speed has one value and at the end another value.

The closest you can come to your flawed statement may be that the average speed during the first second is 5m/s. Maybe you have a closer look at these textbooks.
 
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Vivan Vatsa said:
It is written in many textbooks I have referred. Every where it is explained in the same way. It tells that though respective of how much the body travelled, it will travel with a velocity of 5 m/s in the last second.
I don't understand why is that so...?
Since this is not true, I highly doubt that it is written in many textbooks. I, like nasu, would like to see the actual reference you are basing this on.
 
Ok @nasu thanks for your help.
I think that's the only way possible to understand.
The book is Aakash module(Target 1).
Well thanks for your explanation. I will think over it &I ask some more doubts.
 
It seems it is a textbook used in India.
Maybe you can post an image of the page where you think it says what you think it says. :)
 
Vivan Vatsa said:
It is written in many textbooks I have referred. Every where it is explained in the same way. It tells that though respective of how much the body travelled, it will travel with a velocity of 5 m/s in the last second.
I don't understand why is that so...?

If you throw something straight up (assuming you throw it fast enough so that it travels upwards for more than a second), then:

In the last second before it reaches its highest point it will travel ##5m## (hence have an average speed of ##5m/s## during this second); and, in the first second of its descent it will also travel ##5m##, with the same average speed during that second.

Quite what the significance of this might be is anyone's guess! Really you should be thinking about motion under gravity as a velocity under a continuous rate of change; not as a sequence of discrete average speeds.
 
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PeroK said:
If you throw something straight up (assuming you throw it fast enough so that it travels upwards for more than a second), then:

In the last second before it reaches its highest point it will travel ##5m## (hence have an average speed of ##5m/s## during this second); and, in the first second of its descent it will also travel ##5m##, with the same average speed during that second.

Quite what the significance of this might be is anyone's guess! Really you should be thinking about motion under gravity as a velocity under a continuous rate of change; not as a sequence of discrete average speeds.
Yeah but mainly UNIFORMLY ACCELERATED MOTION UNDER GRAVITY is a game of continuous rate of change working under gravity, Right?
If that is true then I should conclude that motion under gravity is a relation of velocity with respect to time working under a constant(known as Gravity).
 
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Vivan Vatsa said:
Yeah but mainly UNIFORMLY ACCELERATED MOTION UNDER GRAVITY is a game of continuous rate of change working under gravity, Right?
If that is true then I should conclude that motion under gravity is a relation of velocity with respect to time working under a constant(known as Gravity).

Gravity near the Earth's surface is an example of uniform acceleration. The point of kinematics is that, whatever the cause of the uniform acceleration, the same kinematics principles, equations and solutions apply.

What's special about gravity is that the acceleration is independent of the mass of the object, so that for all objects in freefall near the Earth's surface:

##\frac{dv_y}{dt} = g##

Where ##v_y## is the vertical component of the velocity of an object. One reason the ##5m/s## "rule" makes no sense is that gravity does not affect the horizontal component of velocity, which remains constant during freefall.
 
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