THERE ARE TWO MOTION : for the upside motion - F= MG +KV^2 Mdv/dt = MG + KV^2 Mdv/ds . ds/dt=Mg + KV^2 Mv /(Mg KV^2) dv = ds [v0 ][o ] MV/(MG +KV^2). dv = [ 0][l ] ds (try the integration if your maths is strong and you'll get -) M[ ln (MG/(MG + KVo^2) )]/2k = L ( L is the max height attained ) ............1 now for downward motion m dv/dt = MG - KV^2 mV / (mg - kv^2) dv = ds integrate both side [o ][v' ] mV/( mg - kv^2) = [0 ][L] ds again try and you'll get m[ ln ( | mg - kv'^2 |/mg ] / 2k = L .......................2 compare 1 and 2 ln (mg / (mg + KV^2) ) = ln (mg -kv^2 /mg)
v' = [ MGKV0^2 / ( MGK + K^2 V0^2) ]
V' = V0/ (1+ KV0^2 /MG ) V' is your answer definetely ... write if you can't understand my solution i might elaborate it further .
WALK NOT AS IF YOU RULE THE WORLD...
BUT AS IF YOU DON'T CARE WHO RULES IT....
MAN WOULD DO NOTHING IF HE WAITED UNTIL HE COULD DO IT SO WELL THAT NO OTHER BEING CAN FIND FAULT WITH IT..................NEWMAN
Moderation Team
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Mdv/dt = MG + KV^2 
[o ] MV/(MG +KV^2). dv = [ 0]
[ MGKV0^2 / ( MGK + K^2 V0^2) ]
