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2015-02-16T15:22:51+05:30

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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
 


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2015-02-17T13:21:50+05:30

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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

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