Question

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 ?

Answer 1

The law of conservation of energy says

$$PE_i=KE_f$$

$$mgh_i=KE_f$$

Hence

$$h_i=\frac{KE_f}{mg}=\frac{400}{2.0*9.8}=20\:\rm m$$