Asked by Sotonye
A person was reported to have fallen 45.0 m from a balcony of an apartment building onto a ventilator
box, crushing it to a depth of 4.10 m (Fortunately, the person walked away with only minor injuries). What
was the magnitude of the acceleration of the person during the time she was in contact with the box?
A. 9.80 m/s2
B. 17.4 m/s2
C. 34.6 m/s2
D. 76.2 m/s2
E. 108 m/s2
box, crushing it to a depth of 4.10 m (Fortunately, the person walked away with only minor injuries). What
was the magnitude of the acceleration of the person during the time she was in contact with the box?
A. 9.80 m/s2
B. 17.4 m/s2
C. 34.6 m/s2
D. 76.2 m/s2
E. 108 m/s2
Answers
Answered by
Damon
fall 45 m:
(1/2) m v^2 = m g h
so
v = sqrt(2 g h) = sqrt(90*9.81)
= 29.7 m/s
Vi = 29.7
average speed during stop = 14.9
so stopping time = 4.10/14.9 = .279 seconds
a = dv/dt = 29.7/.279 = 106 m/s^2
looks like E, what are you using for g I wonder
(1/2) m v^2 = m g h
so
v = sqrt(2 g h) = sqrt(90*9.81)
= 29.7 m/s
Vi = 29.7
average speed during stop = 14.9
so stopping time = 4.10/14.9 = .279 seconds
a = dv/dt = 29.7/.279 = 106 m/s^2
looks like E, what are you using for g I wonder
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