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Mathematics 22 Online
OpenStudy (anonymous):

.

OpenStudy (anonymous):

A stone is thrown vertically upward with an initial speed u from the top of a tower, reaches the ground with a speed 3u. The height of the tower

OpenStudy (vishweshshrimali5):

Try using conservation of mechanical energy

OpenStudy (anonymous):

We can use v^2 = u^2 +2as

OpenStudy (vishweshshrimali5):

Yeah sure

OpenStudy (vishweshshrimali5):

@No.name please please post the questions in relevant section :)

OpenStudy (anonymous):

how to use , i have a problem considering signs @vishweshshrimali5 , i knew u would say this , but this was in math

OpenStudy (vishweshshrimali5):

Really ?

OpenStudy (vishweshshrimali5):

Well okay for signs

OpenStudy (vishweshshrimali5):

Consider a reference coordinate system

OpenStudy (vishweshshrimali5):

|dw:1405487303957:dw|

OpenStudy (vishweshshrimali5):

Like this ^^^

OpenStudy (anonymous):

Yes , in dynamics, i will give u a screen shot of that later , i gave ganesh

OpenStudy (aum):

\(\Large mgh + \frac 12mu^2 = \frac 12m(3u)^2\)

OpenStudy (anonymous):

OK

OpenStudy (vishweshshrimali5):

Then your u would be +ve, 3u would be negative, g would be negative

OpenStudy (vishweshshrimali5):

Or you can use the conservation mechanical energy as @aum used :) (I generally use this one as it doesn't uses signs)

OpenStudy (anonymous):

Wait 5u^/g ?

OpenStudy (vishweshshrimali5):

?

OpenStudy (anonymous):

5u^2/g

OpenStudy (vishweshshrimali5):

Sorry would have to go now :(

OpenStudy (anonymous):

oh @Abhisar then is it right

OpenStudy (anonymous):

@aum am i correct

OpenStudy (abhisar):

u got it @No.name ?

OpenStudy (anonymous):

i got the answer as 5u^2/g is it correct

OpenStudy (abhisar):

okay..let's see !

OpenStudy (anonymous):

Okay

OpenStudy (abhisar):

What will be the final velocity when the stone returns at the top of the tower ?

OpenStudy (anonymous):

Yes wait that's were the problem is would it be (-3u) or -(3u) ?

OpenStudy (anonymous):

why - u It is thrown upwards

OpenStudy (abhisar):

|dw:1405487773484:dw|

OpenStudy (abhisar):

Its okay.... now use third equation of motion to find s

OpenStudy (aum):

By both methods I am getting \(\Large h = \frac{4u^2}{g}\)

OpenStudy (abhisar):

and that is correct !

OpenStudy (anonymous):

(-3u)^2 = u^2 -2gs

OpenStudy (abhisar):

\(\huge\checkmark\)

OpenStudy (aum):

+2gs

OpenStudy (abhisar):

+2gs

OpenStudy (abhisar):

yep !

OpenStudy (anonymous):

g is negative right?

OpenStudy (abhisar):

No, if u take 3u +ve then g must be +ve

OpenStudy (abhisar):

if u take g -ve then 3u will also be -ve

OpenStudy (anonymous):

I take 3u as negative

OpenStudy (aum):

If you take the downward direction as positive, then u, 3u, g are all positive.

OpenStudy (abhisar):

yes that's what actually happens !

OpenStudy (anonymous):

Then u would be negative because the ball is thrown UPwards

OpenStudy (abhisar):

It's all on the basis of cartesian coordinates.

OpenStudy (aum):

Not talking about the upward u but the u when it is traveling downwards

OpenStudy (anonymous):

what??

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