A body is thrown vertically up with a velocity of 100m//s and another one is thrown 4 sec after the first one. How long after the first one is thrown will they meet?

A body is thrown vertically up with a velocity of 100m//s and another

1.	A body is thrown vertically up with a velocity of 100m//s and another one is thrown 4 sec after the first one. How long after the first one is thrown will they meet?
Answer» Solution :Let them MEET after t sec. `S_(1)=100t-(1)/(2)g t^(2)andS_(2)=100(t-4)-(1)/(2)g(t-4)^(2)` `therefore 100t-(1)/(2)g t^(2)=100(t-4)-(1)/(2)g(t-4)^(2)` `therefore 400=(1)/(2)g[t^(2)-(t-4)^(2)]=(1)/(2)g.4(2t-4)` `therefore 2t-4=(800)/(4g)=20," if "g=10m//s^(2) therefore t=12sec`