Just before striking the ground, a 2.0-kg mass has 400 J of KE. If friction can be ignored, from what height was it dropped ?
The law of conservation of energy says
"PE_i=KE_f""mgh_i=KE_f"
Hence
"h_i=\\frac{KE_f}{mg}=\\frac{400}{2.0*9.8}=20\\:\\rm m"
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