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we see the first equation gives
t
=
x
/
v
0
cos
θ
0
, and when this is substituted into the second
the result is
yx
gx
v
=−
tan
cos
.
0
2
0
22
0
2
One may solve this by trial and error: systematically trying values of
0
until you find the
two that satisfy the equation. A little manipulation, however, will give an algebraic
solution: Using the trigonometric identity 1 / cos
2
0
= 1 + tan
2
0
, we obtain
1
2
1
2
0
2
0
2
2
00
2
0
2
gx
v
xy
gx
v
tan
tan
θθ
−+
+
=
which is a secondorder equation for tan
0
. To simplify writing the solution, we denote
()
(
)
2
11
0
/
9.80 m/s
50 m
/ 25 m/s
19.6m.
cg
x
v
==
=
Then the secondorder equation becomes
c
tan
2
0
–
x
tan
0
+
y
+
c
= 0.
Using the
quadratic formula, we obtain its solution(s).
(
)
0
4
50 m
(50 m)
4 3.44 m 19.6 m 19.6 m
tan
.
2
2 19.6 m
xx
y
c
c
c
±−+
±
−
+
The two solutions are given by tan
0
= 1.95 and tan
0
= 0.605. The corresponding (first
quadrant) angles are
0
= 63° and
0
= 31°. Thus,
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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