i didnt get wich is the surface?
any way
while going up , net force = - (mg+cu2) ( j )
and while going down net force = (cu2 - mg ) ( j )
now while going up ,
accn a = mg+ cu2m = - dudt= - ududx
so
mudumg+cu2 = -dx
i.e mclog(mg+cu2)= -x
where u changes from uo to u , and x changes from 0 to x
(x measured from lowest point )
so, - cx = mlog(mg+cu2mg+cuo2)
i m assuming that the surface is the one frm which the particle is thrown
while going down
accn a = mg-cu2m = +dudt= + ududx
mudumg-cu2 = dx
i.e
- mclog(mg-cu2) = x
where u changes from 0 to some u
and x changes from 0 to some x (measured from highest point )
so - mclog(mg-cu2mg) = x
so we got 2 eqns
- x = mclog(mg+cu2mg+cuo2) ........(1)
- mclog(mg-cu2mg) = x ...........(2)
in eqn 1 , put u = 0 , and x =xmax
in eqn 2 , u is the u required , and put x = xmax
and equate xmax in the 2 eqns
so mgmg+cuo2 = mg-cu2mg
m2g2 = m2g2 +cmg(uo2-u2) - c2uo2u2
so mg(uo2-u2)=cuo2u2, i.e mguo2=u2(cuo2+mg)
u2 = mguo2cuo2+mg
wats the ans given ?