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Let the both meet at time t A covers distance x in time t and B covers distance 100-x in time t let u be velocity of A with which it is projected so when A is at 100 m its velocity is zero so -u²=2(-10)(100) u=10√20 so 100-x=1/2(10)t²=5t² ...(1) for B and x=10√20t-5t²....(2) add (1)+ (2) 100=10√20t t=10/√20=√(10/2)=√5s they meet after √5 secs 100-x=5(5)=25 x=100-25=75 so they meet at 75 m ground