fisica1-youn-e-freedman-exercicios-resolvidos

2
2.86: a) The helicopter accelerates from rest for 10.0 s at a constant 5.0 m s . It thus
reaches an upward velocity of
2
v = v + a t = (5.0 m s )(10.0 s) 50.0 m s
and a height of
y
0 y y
=
1 2 1
2
2
y = a
yt
= (5.0 m s )(10.0 s) = 250 m at the moment the engine is shut
2
off. To find the helicopter's maximum height use
2 2
v = v + a ( y − y )
Taking
y
2
0 y
2
y 0
y = 250 m
0
, where the engine shut off, and since v 2 = y
0 at the maximum height:
y
max
− y
0
− v
=
2g
2
(50.0 m s)
ymax
= 250 m −
= 378 m
2
2( −9.8
m s )
or 380 m to the given precision.
2
0 y
b) The time for the helicopter to crash from the height of 250 m where Powers
stepped out and the engine shut off can be found from:
1 2
2
1
2
y = y0 + v0
y
t + ayt
= 250 m + (50.0 m s) t + ( −9.8 m s ) t = 0
2
2
where we now take the ground as y = 0 . The quadratic formula gives solutions of
t = 3.67s and 13.88 s, of which the first is physically impossible in this situation. Powers'
position 7.0 seconds after the engine shutoff is given by:
1
2
2
y = 250 m + (50.0 m s)(7.0 s) + ( −9.8 m s )(49.0 s ) = 359.9 m
2
at which time his velocity is
2
v = v + gt = 50.0 m s + ( −9.80 m s )(7.0 s) = 18.6 m s
y
0 y
−
Powers thus has 13 .88 − 7.0 = 6.88s
more time to fall before the helicopter crashes, at his
2
constant downward acceleration of 2.0 m s . His position at crash time is thus:
1 2
y = y0
+ v0
y
t + ayt
2
1
2
= 359.9 m + ( −18.6
m s)(6.88s) + ( −2.0 m s )(6.88s)
2
= 184.6 m
or 180 m to the given precision.
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