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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
Saurabh 1
When the given object is at some height ( let that height be "h" ). then, by law of conservation of energy decrease in potential energy (due to height) = increment in kinetic energy (due to acquired velocity). mgh= 1/2(mv²) , where v is the velocity acquired. clearly m gets cancelled from both side, so final equation for velocity "v" is v² = 2gh v= √(2gh) Just put h= 1m and g= 10m/s² then, v= √(2×10×1) = √20 v= √20 = 4.47m/s So upto one decimal place, speed with which it will hit the ground (v) = 4.5 (in m/s). So final answer is -- 4.5