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Physics 125
Discussion #4 Solution
1
Problem 1: A body projected horizontally
A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is
horizontal, with magnitude 9.0 m/s.
x
direction
y
direction
initial position
x
0
= 0
y
0
= 0
initial velocity
v
0x
= +9.0 m/s
v
0y
= 0
acceleration
a
x
= 0
a
y
=  9.8 m/s
2
(3)
What is the motorcycle’s position (
x
,
y
) at t = 0.50 s?
(m)
2
.
1
)
50
.
0
)(
8
.
9
(
2
1
0
2
1
(m)
5
.
4
)
50
.
0
)(
0
.
9
(
2
2
0
0

=

+
=
+
=
=
=
=
at
t
v
y
t
v
x
y
x
(1)
First, let’s define our coordinate system. The
coordinate system is shown in the figure. We
choose the origin at the edge of the cliff, where
the motorcycle first becomes a projectile.
(2)
List all the known factors (in both x and y
directions).
0
v
h
x
y
O
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Discussion #4 Solution
2
(4)
What is the motorcycle’s distance from the origin at t = 0.50 s?
(m)
7
.
4
)
2
.
1
(
)
5
.
4
(
2
2
2
2
=

+
=
+
=
y
x
r
(5)
What are the x and y components of velocity at t = 0.50 s?
(m/s)
9
.
4
)
50
.
0
)(
8
.
9
(
0
(m/s)
0
.
9
0
0

=

+
=
+
=
+
=
=
at
v
v
v
v
y
y
x
x
The motorcycle has the same horizontal velocity v
x
as when it left the cliff
at t=0, but in addition there is a vertical velocity v
y
directed downward (in
the negative ydirection.
(6)
Express the velocity at t = 0.50 s in terms of magnitude and direction.
o
x
y
y
x
v
v
v
v
v
29
0
.
9
9
.
4
tan
tan
(m/s)
2
.
10
)
9
.
4
(
)
0
.
9
(
1

1

2
2
2
2

=

=
=
=

+
=
+
=
α
At t = 0.50 s the velocity is 29
o
below
the horizontal.
v
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 Fall '08
 unknown
 Projectile Motion

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