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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 Jan 2008 23:11:14 IST
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u = (2g{H-h})^1/2
v = (2gH)^1/2
v = u + gt
t = v - u/g
t = g^1/2 [ (2H)^1/2 - (2 {H-h})/g
= (2H)^1/2 - (2{H-h})^1/2/ (G)^1/2
Since horizontal dist to be covered = x
so horizontal velocity = x / (2H)^1/2 - (2{H-h})^1/2/ (G)^1/2
= x (g)^1/2 / [ (2H)^1/2 - (2{H-h})^1/2]
now by principal of conservation of momentum.....
m * V reqd = M * x (g)^1/2 / [ (2H)^1/2 - (2{H-h})^1/2]
time reqd by the bag to cover the dist....
= M * x (g)^1/2 / m[ (2H)^1/2 - (2{H-h})^1/2] * [ (2H)^1/2 - (2{H-h})^1/2] / (g)^1/2
= Mx/m
please nudge if not clear and sorry for not being able to post a picture
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All that is gold does not glitter
Not all who wander are lost
The old that is strong does not wither,
Deep roots are not reached by frost.
From ashes a fire shall be woken
From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king. |
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