A particle is projected vertically up with velocity v=(4gRe/3)1/2 from earth surface. The velocity of particle at height equal to half of the maximum height reached by it .....?

Determination of maximum height:

Potential energy + Kinetic energy at the surface = Potential energy at the maximum height

-GMmRe+12mv2=-GMmRe+h-GMmRe+12m4gRe3=-GMmRe+h-GMmRe+4m6GMRe=-GMmRe+h-1Re+23Re=-1Re+h13Re=1Re+h        h=2Re

Total energy at the surface = Total energy at the half of maximum height:

-GMmRe+12m4gRe3=12mv'2-GMmRe+h2-GMmRe+4m6GMRe=12mv'2-GMm2Re-GMm2Re+2GMm3Re=12mv'2GMm6Re=12mv'2GM6Re=12v'2GM3Re=v'2     v'=GM3ReBut, g'=GMRe+h22            =GM4Re2    4g'Re=GMReHence,      v'=4g'Re3      g=GMRe2   g'g=Re24Re2        =g4So,v'=4g'Re3   =4g4Re3   =gRe3      

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