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Utter confusion regarding distance covered during acceleration in free fall.

by AakashPandita
Tags: acceleration, confusion, covered, distance, fall, free, utter
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AakashPandita
#1
Nov16-12, 11:14 AM
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P: 133
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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rcgldr
#2
Nov16-12, 11:42 AM
HW Helper
P: 7,126
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?
AakashPandita
#3
Nov17-12, 09:54 AM
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P: 133
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?

rcgldr
#4
Nov17-12, 02:13 PM
HW Helper
P: 7,126
Utter confusion regarding distance covered during acceleration in free fall.

Quote Quote by AakashPandita View Post
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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