Prove that total mechanical energy remains constant for a mass M dropped from a tower of height H
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when body is at rest the u be 0 and k.e
will be 0 p.e=mgh k+u=0+mgh
mgh
ii)let body be moving with velocity v distance s u be0 and gravity be g
from formula
v2=u2+2as
v=0+2gs
k.e.=1/2mv2
1/2*2gs
p.e.=mgh
mgh
iii) when body has reached ground
again u will be 0 as it is in rest from formula v2=u2+2as
v2=0 +2gh
k.e. 1/2mv2
substitute your values and you will get mgh
thus k+ u remains constant at every position
will be 0 p.e=mgh k+u=0+mgh
mgh
ii)let body be moving with velocity v distance s u be0 and gravity be g
from formula
v2=u2+2as
v=0+2gs
k.e.=1/2mv2
1/2*2gs
p.e.=mgh
mgh
iii) when body has reached ground
again u will be 0 as it is in rest from formula v2=u2+2as
v2=0 +2gh
k.e. 1/2mv2
substitute your values and you will get mgh
thus k+ u remains constant at every position
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