(25%) Problem 4: Standing at the base of one of the cliffs of Mt. Arapiles in Vi
ID: 1771698 • Letter: #
Question
(25%) Problem 4: Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 95 m. He can't see the rock right away but then does, 1.2 s later. 50% Part (a) How far above the ground, in meters, is the rock when the hiker starts to see it? Neglect the time for the sound to reach the hiker Grade Summary Deductions Potential 0% 100% sin cosO cotan0 asinO acos0 tan() | | ( 78 9 Submissions Attempts remaining: 10 % per attempt) detailed view atan() acotanO s acotanO S sinh0 0 coshOtanh0 cotanh0 Degrees O Radians END BACKSPACE DEL CLEAR Submit Hint I give up! Hints: 3% deduction per hint. Hi ints remaining: Feedback: 3% deduction per feedback. D 50% Part (b) How much time, in seconds after he notices the falling rock, does he have to move before the rock hits his head? Ignore the height of the hiker.Explanation / Answer
(a)
y = V0t+1/2 at^2 = 1/2*(9.8 m/s^2)*(1.2s)^2 = 7.05m
so the rock fall 7.05 m in the 1.2s befor the hiker sees the rock when
he finally sees the rock it is 87.95 m above his head
(b)
y = 1/2 at^2 = t = 2y/a = 2*(95m)/9.80m/s^2 = 4.40s so the rock will
take 4.40s to fall the full distance
thus the hiker will have 4.40s-1.2s = 3.2s to move out of the way before the
rock strikes the hiker location ignoring the height of person
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