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Mechanics help!!

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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 ---
mm, sir i am confused, why have you done it this way??, wont the maximum height= 0.36 m ( which it was initially at )+ the total distance partical A will cover while falling which is 0.36 m aswell ///

thanks for the help with the other questions :)

astarmathsandphysics:
Wait

astarmathsandphysics:
o.36+extra distance travelled by B after A hits the ground - during this time the acceleration chages from 2m/s^2 to -9.8m/s^2

mousa:

--- Quote from: astarmathsandphysics on February 06, 2010, 12:25:09 pm ---o.36+extra distance travelled by B after A hits the ground - during this time the acceleration chages from 2m/s^2 to -9.8m/s^2

--- End quote ---

why??? ::) does the acceleration change??

astarmathsandphysics:
when falling freely under gravity the acceleration is always -9.8

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