Mathematics
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OpenStudy (anonymous):
.
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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 :)
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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|
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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
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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 :(
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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 !
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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|
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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):
\(\huge\checkmark\)
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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
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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
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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??