B Uniformly accelerated motion under gravity

Hello Vivan,

This sounds very unbelievable to me. Can you specify where you picked this up and how exactly it was formulated ?

 

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.

 

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.

 

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. :)

 

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.

 

Gravity near the Earth's surface is an example of uniform acceleration. The point of kinematics is that, whatever the cause of the uniform accleration, 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.