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2015-03-10T15:54:00+05:30

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Just before reaching the ground 
h=0 m 
so potential energy = 0
and initial velocity = 0 so 
2gh=v^2
2*10*10=v^2
200=v^2
v=√200m/s
now k.e = 1/2*2*√200*√200
           =200 J
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2015-03-10T15:58:08+05:30

This Is a Certified Answer

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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.
Mass of body, m = 2kg

it is lifted to height = 10m
initial KE = 0
initial PE = mgh = 2×10×10 = 200J

when it is dropped, just before touching ground,
height = 0
final PE = mgh = 2×10×0 = 0J

According to conservation of energy,
Total initial energy = Total final energy
⇒ 0 + 200 = 0 + final KE
⇒ final KE = 200J

So, just before touching ground, KE is 200J and PE is 0J.
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