HomeWork 3 solutions
Physics 203
1.
Cutnell & Johnson 8
th
Edition, Chapter 2, Problem 44
REASONING
Because there is no effect due to air resistance, the rock is in free fall from its launch until it
hits the ground, so that the acceleration of the rock is always
−9.8
m/s
2
, assuming upward to be the positive
direction. In (a), we will consider the interval beginning at launch and ending 2.0 s later. In (b), we will
consider the interval beginning at launch and ending 5.0 s later. Since the displacement isn’t required,
Equation 2.4
( )
0
v
v
at
=
+
suffices to solve both parts of the problem. The stone slows down as it rises, so
we expect the speed in (a) to be larger than 15 m/s.
The speed in (b) could be smaller than 15 m/s (the rock
does not reach its maximum height) or larger than 15 m/s (the rock reaches its maximum height and falls
back down below its height at the 2.0s point).
SOLUTION
a.
For the interval from launch to
t
= 2.0 s, the final velocity is
v
= 15 m/s, the acceleration is
a
=
−9.8
m/s
2
, and the initial velocity is to be found.
Solving Equation 2.4
( )
0
v
v
at
=
+
for
v
0
gives
2
0
15 m/s ( 9.8 m/s )(2.0 s) 35 m/s
v
v at
= − =
−−
=
Therefore, at launch,
Speed
35 m/s
=
b.
Now we consider the interval from launch to
t
= 5.0 s.
The initial velocity is that found in part (a),
v
0
= 35 m/s.
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 Fall '08
 RICHARDSON, B
 Physics, Resistance, Acceleration, Force, Friction, Mass, Work, Velocity

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