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Physics 19 Online
OpenStudy (anonymous):

tuti drops a marble into a 20 m deep well g=10 m/s^2. what is the time needed by the marble to reach the surface of water in the well?

OpenStudy (abhisar):

u=0 s=20 a=10 use the same formula we used in the last question and find the value of t

OpenStudy (anonymous):

I thought using the vt^2=2gh formula for this one, am i mistaken? physics is a bit confusing, and there are no proper information in my module

OpenStudy (abhisar):

But you don't know the value of v, so u can't use this formula here.

OpenStudy (anonymous):

Right... Thanks..

OpenStudy (abhisar):

use \(\sf huge s=ut+\frac{1}{2}at^2\) answer will be 4seconds

OpenStudy (anonymous):

I got 5 seconds

OpenStudy (abhisar):

20=0*t+0.5*10*\(\sf t^2\) =>\(\sf t^2\)=4 =>t=2 seconds

OpenStudy (abhisar):

Sorry answer will be 2 seconds

OpenStudy (anonymous):

okay thanks

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