(a) Find the free-fall acceleration on its surface. (b) A cliff on Miranda is 5.
ID: 1517464 • Letter: #
Question
(a) Find the free-fall acceleration on its surface.
(b) A cliff on Miranda is 5.00 km high. It appears on the limb at the 11 o'clock position in part a of the figure above and is magnified in part b of the figure above. A devotee of extreme sports runs horizontally off the top of the cliff at 5.50 m/s. For what time interval is he in flight? (Ignore the difference in g between the lip and base of the cliff.)
(c) How far from the base of the vertical cliff does he strike the icy surface of Miranda?
(d) What is his vector impact velocity?
Explanation / Answer
(a) Find the free-fall acceleration on its surface.
From the given data
g = Gm/r^2 = 6.67x10^-11 x 6.68 x 10^19 / (2.42 x 10^5)^2
g = 0.076m/s^2
(b) A cliff on Miranda is 5.00 km high. It appears on the limb at the 11 o'clock position in part a of the figure above and is magnified in part b of the figure above. A devotee of extreme sports runs horizontally off the top of the cliff at 5.50 m/s. For what time interval is he in flight?
s = ut + 1/2at^2
5000 = 0 + 1/2 x 0.076 x t^2
t = 362 secs
(c) How far from the base of the vertical cliff does he strike the icy surface of Miranda?
s = v x t = 5.5 x 362 = 1991 m
d) What is his vector impact velocity?
vertical velocity on impact = at = 0.076 x 362 = 27.5m/s
vector velocity^2 = Vv^2 + Vh^2
Vv^2 = 27.5^2 + 5.5^2
Vv^2 = 786.5
Vv = 28.044 m/s
Tan(theta) = 27.5 / 5.5 = 5
angle = 79 deg below horizontal.
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