Question

an object of mass 10kg is dropped from a height of 100 cm. find:
a) its kinetic energy
b)velocity just as it reaches the ground (g=10m/s2)

Answer 1

mass of the object ( m ) = 10 kg
  height (h) = 100cm = 1m
  g = 10m/s
K.E.= $\frac{1}{2} mv^2$
  
when the body reaches the ground its P.E. changes into K.E.
object on the top = K.E.on the ground 
 mgh = K.E.on the ground 
 gh= $\frac{1}{2} v^2$
 v = $\sqrt{2gh}$
v = $\sqrt{2*10*1}$
v =  $\sqrt{20}$ m/s

 P.E = mgh
     = 10 * 10 * 1
     = 20 j
 

Answer 2

Final velocity of the body when it reaches the ground (100 cm =1 m)

v^2 = u^2 - 2as = 0 - {2*(-10)*1} = 20

So velocity just as it reaches the ground = (Square root of 20) m/s

Kinetic energy = 1/2*m*v^2 = 1/2*10*20 = 100 J