Ball a is thrown up vertically with speed 10m/s.at the same instant another ball b is released from rest at height h . At time t the speed of a relative to b is

							
At time t,
speed of a= 10 – gt
 
speed of b = 0 – gt
 
Speed of a relative to b = 10 – gt + gt = 10m/s
						

							speed of a,v1=10-gtspeed of b,v2=0-gt                        =-gtspeed of a relative to b=v1+(-v2)                                    Vab=10-gt+(-(-gt))                                    Vab=10-gt+gt                                    Vab=10m/s
						

							But why the manitude of v2 is 0-gt? In this case ball b is in downward direction.  So g is +ve.  Please tell me as early as possible.  I can not understand this in second case of your answer.
						

							because here the direction is seen wrt ball a i.e. upward direction is positive. so even though ball b is accelerating it will be taken as -g
						

							
V of a= 10+gt. ( Taking upward direction as +ve)
V of b= 0+(-g)t = -gt
We know that relative v = v of a- v of b
But here both the velocities are opposite to each other in directions so take one of them as -ve ( here taking v of b as -ve)
So relative velocity= v of a -(-vof b)
= 10+gt +(-gt)
=10m/s