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41. Following the hint, we have the timereversed problem with the ball thrown from the
ground, towards the right, at 60° measured counterclockwise from a rightward axis. We
see in this timereversed situation that it is convenient to use the familiar coordinate
system with +
x
as
rightward
and with positive angles measured counterclockwise.
(a) The
x
equation (with
x
0
= 0 and
x
= 25.0 m) leads to
25.0 m = (
v
0
cos 60.0°)(1.50 s),
so that
v
0
= 33.3 m/s.
And with
y
0
= 0, and
y
=
h
> 0 at
t
= 1.50 s, we have
yy v
t g
t
y
−=
−
00
1
2
2
where
v
0
y
=
v
0
sin 60.0°.
This leads to
h
= 32.3 m.
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 Spring '08
 Any
 Physics

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