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52. For
Δ
y
= 0, Eq. 422 leads to
t
= 2v
o
sin
θ
o
/
g
, which immediately implies
t
max
= 2
v
o
/
g
(which occurs for the “straight up” case:
o
= 90
°
). Thus,
1
2
t
max
=
v
o
/
g
¡
1
2
= sin
o
.
Therefore, the halfmaximumtime flight is at angle
o
= 30.0
°
.
Since the least speed
occurs at the top of the trajectory, which is where the velocity is simply the
x
component
of the initial velocity (
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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, Light

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