A particle is thrown with a velocity of u m/s.  It passes A and B as shown in figure at time  $t_1=1$s and  $t_2=3$s' The value of u is      

 

If the projectile takes 1 sec to reach A it will take same time 1 sec to come down from B as they are in the same height. So total time of flight is time to reach B and then reach the ground. i.e. 3+1=4 sec.

$$\begin{gathered} T=\frac{2u\sin \theta }{g}=4 \\ u=\frac{4g}{2\sin \theta }=40 m/s \left(\theta ={30}^0\right) \end{gathered}$$