A swimmer dived off a cliff with a running horizontal leap. What musthis minimum speed be just as he leaves the top of the cliff so that hewill miss the ledge at the bottom which is 2 m wide and 9 m below thetop of the cliff?

time taken by swimmer to reach the ground is t

       s=uyt+at^2/2             (uy is verticle component of velocity of swimmer which is zero)

       s=9 and a=g=10

       so t=square root(2s/g)

             =3/root5 secs

  horizontal distance covered during this time is t.v (v is the velocity in horizontal direction)

    horizontal distance should be  2m so that he will mis the ledge

         vt=2

         v=2/t

           =2.root5/3 m/s

Approved