If a body of mass 3 kg is dropped from the top of a tower of height 25m. Then its kinetic energy after 3 sec is .. don't give any links. pls solve

Given that , m = 3kg , u = 0 , t = 3 sec , a = g = 9.8m/s² (acc due to gravity)
So velocity at t = 3sec
v = u +a t 
   = 0 + 9.8(3)
  = 29.4 m/s
Kinetic energy = 1/2 m v²
                        = 1/2 (3) (29.4)²
                       = 1296.54 J
Hope this helps...