Two boys are standing at the ends A and B of a ground, where `AB=a`. The boy at B starts running in a direction perpendicular to AB with velocity `v_(1)`. The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is :

Explanation

Let two boys meet at point C after time 't' from the starting.
Then AC=vt, BC`=v_(1)t` but `(AC)^(2)=(AB)^(2)+(BC)^(2)rArrv^(2)t^(2)=a^(2)+v_(1)^(2)t^(2)`
`rArrt^(2)(v^(2)-v_(1)^(2))=a^(2)rArrt=sqrt((a^(2))/(v^(2)-v_(1)^(2)))`