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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
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