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position for this part of the problem) and take +
x
rightward and +
y
downward. Then
(since the initial velocity is purely horizontal) the projectile motion equations become
xv
t
y
g
t
==
−
and
1
2
2
.
Solving for
x
at the time when
y = h
, the second equation gives
th
g
=
2.
Then,
substituting this into the first equation, we find
()
2
2 2.0 m
2
7.48 m/s
4.8 m.
9.8 m/s
h
g
=
11. To find where the ball lands, we need to know its speed as it leaves the track (using
conservation of energy). Its initial kinetic energy is
K
i
= 0 and its initial potential energy
is
U
i
= M gH
. Its final kinetic energy (as it leaves the track) is
KM
v
I
f
=+
1
2
2
1
2
2
ω
(Eq.
11-5) and its final potential energy is
M gh
. Here we use
v
to denote the speed of its

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