Two boys are standing at the end 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 simultanieously with velocity V and catches the other boy in a timet, where t is 

Velocity of A relative to B is given by

$V_{AB}=\overline{V_A}-\overline{V_B}=\overline{V}-\overline{V_1}\to \left(1\right)$ by taking x-components of eq(1),  we get 
$O=v\sin \theta -v_1\Rightarrow \sin \theta =\frac{v_1}{v}\to \left(2\right)$
By taking y-components of eq(1), we get  $V_y=V\cos \theta \to \left(3\right)$
Time taken by boy at A to catch the boy B is 
$t=\frac{relatives}{relatives}\frac{displacements}{velocity}\frac{along}{along}\frac{y-axis}{y-axis}=\frac{a}{v\cos \theta }=\frac{a}{v\sqrt{1-{\sin }^2\theta }}$
$=\frac{a}{v\sqrt{1-{\left(\frac{v_1}{v}\right)}^2}}=\sqrt{\frac{a^2}{v^2-v_1^2}}$