Utter confusion regarding distance covered during acceleration in free fall.

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In a gravitational field with an acceleration of 10 meters per second squared, an object falling for one second reaches a final velocity of 10 meters per second. The distance covered during this time is calculated as 10 meters, which raises questions about average speed since the object starts from rest. The average speed during the fall is actually 5 meters per second, not 10 meters per second, due to the constant acceleration. When considering a free fall from rest over a distance of 10 meters, the final velocity is determined to be approximately 14.14 meters per second, calculated as the square root of 200. Understanding these concepts clarifies the relationship between distance, speed, and acceleration in free fall scenarios.
AakashPandita
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Let there be an obj. which is made to fall from a certain height in a gravitaitonal field.
where

a= 10 metre per second squared

Say it falls for 1 second.

v(final velocity)= 10 metre per second.

How much distance does it cover in that 1 second?

distance = speed x time = avg. speed x 1second = 10 m

This means that 10 m is covered during that 1 second.

But how could that be when the velocity had not reached 10 metre per second?
 
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If acceleration is constant, and if the starting speed is 0 meters per second and the final speed is 10 meters per second, then what is the average speed?
 
thank you. i understood my mistake.
But now i have another doubt.

A body falls freely for 10 metres from rest.

what is the final velocity?

i found it is sqrt200.
am i right?

i found sqrt200.
am i right?
 
AakashPandita said:
A body falls freely for 10 metres from rest. what is the final velocity? i found it is sqrt200.
Correct. You can also write this as 10 x sqrt(2) or ~14.1421. I'm not sure what form of answer your class would prefer.
 
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