(a) What potential energy does the rock possess relative to the base of the cliff?
J
(b) The rock falls from the cliff. What is its kinetic energy just before it strikes the ground?
J
(c) What speed does the rock have as it strikes the ground?
m/s
A 17 kg rock, m, is on the edge of a 110 m cliff...?
2015-12-09 6:04 am
回答 (1)
2015-12-14 7:58 am
(a) PE = m*g*h = 17*9.8*110 = 18326 J
(b) KE = PE = 18326 J
(c) v = √(2gh) = √(2*9.8*110) = 46.4 m/s
(b) KE = PE = 18326 J
(c) v = √(2gh) = √(2*9.8*110) = 46.4 m/s
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