A 20-kg rock is on the edge of a 100 meter cliff, as shown in the picture link.
a. What potential energy does the rock possess relative to the base of the cliff?
b. THe rock falls from the cliff. What is its kinetic energy just before it strikes the ground?
c. What speed does the rock have as it srikes the ground?
*Picture link: http://img95.imageshack.us/img95/2082/scan0002zd3.jpg
Please help me with this problem.
I'm having trouble solving it.
2007-01-26 13:50:13 · 4 answers · asked by swimmertommy 1 in Science & Mathematics ➔ Physics
You have put about 5 times as much effort into trying to get an answer by devious means than it takes to get the answer by learning the material.
2007-01-26 14:54:28 · answer #1 · answered by Helmut 7 · 1⤊ 2⤋
P.E. = mgh (m = mass; g = gravity 9.81m/s²; h = height in metres).
a). P.E. = 20 x 9.81 x 100 = 19620 joules = 19.620 kJ
b). At the point just before impact, K.E. = P.E. = 19.620 kJ
c). KE = ½mv²
v² = KE / (½m) = 19.62 / 10 = 1.96
Velocity = √1960 = 44.27 metres/sec
2007-01-26 14:55:46 · answer #2 · answered by Norrie 7 · 2⤊ 0⤋
a: mgh
b 1/2mv^2 = mgh
2007-01-26 13:56:54 · answer #3 · answered by arbiter007 6 · 0⤊ 0⤋
19,600 -apex
2015-11-13 01:47:08 · answer #4 · answered by Kayla'Nycole 1 · 0⤊ 0⤋