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