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Mechanics help!!

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mousa:

--- Quote from: astarmathsandphysics on February 06, 2010, 01:44:25 pm ---when falling freely under gravity the acceleration is always -9.8

--- End quote ---

so what exactly is the 2 m/s^2 that we found?? and about which particle are you talking about?

mousa:

--- Quote from: astarmathsandphysics on February 06, 2010, 09:25:40 am ---6i)s=1/at^2 so a=2s/t^2=2m/s^2
ii)v=u+at=2*9.6=1.2m/s
s=
u=1.2
v=0
a=-9.8
t=
v^2=u^2+2as
0=1.44-19.6s so s=1.44/19.5


--- End quote ---

one thing else, why did you take acce as 2 m/s2 while finding v= 1.2 and then took it as -9.8?? i am confuseeeeeeeeeeeeeed  :P

astarmathsandphysics:
There are tow SEPARATE INSTANCES
before A hits the foor a=2 when a hits the floor v=1.2
After A hit the floor the final velocity of A becomes the initial velocity of B and a becomes -9.8

WHENEVER YOU HAVE A COLLISION YOU HAVE TO USE SUVAT ALL OVER AGAIN

mousa:

--- Quote from: astarmathsandphysics on February 06, 2010, 03:05:52 pm ---There are tow SEPARATE INSTANCES
before A hits the foor a=2 when a hits the floor v=1.2
After A hit the floor the final velocity of A becomes the initial velocity of B and a becomes -9.8

WHENEVER YOU HAVE A COLLISION YOU HAVE TO USE SUVAT ALL OVER AGAIN

--- End quote ---

Thank you sir!!, i finally got it!!

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