A particle is dropped from height H above the ground. After 3 seconds the particle is stopped and again dropped and it takes further 4 seconds to reach the ground. The total time of fall if the particle were not stopped would be

The particle travels distance of $\frac{1}{2}g{\left(3\right)}^2$in first 3 seconds, and it is stopped ,

 So in further 4s it travels distance of $H-\frac{1}{2}g{\left(3\right)}^2$and it is equal to$\frac{1}{2}g{\left(4\right)}^2$

$H-\frac{1}{2}g{\left(3\right)}^2=\frac{1}{2}g{\left(4\right)}^2$……. (1)

If the particle was not sopped it travel distance H, let time of fall be t,

$H=\frac{1}{2}g{\left(t\right)}^2$……… (2)

From 1 and 2 we get t=5s