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43. We neglect air resistance for the duration of the motion (between “launching” and “landing”), so
a
=
−
g
=
−
9
.
8m/s
2
(we take downward to be the
−
y
direction). We use the equations in Table 2-1 (with
∆
y
replacing ∆
x
) because this is
a
= constant motion.
(a) At the highest point the velocity of the ball vanishes. Taking
y
0
=0,weset
v
=0in
v
2
=
v
2
0
−
2
gy
,
and solve for the initial velocity:
v
0
=
√
2
.S
in
c
e
y
= 50 m we Fnd
v
0
=31m/s.
(b) It will be in the air from the time it leaves the ground until the time it returns to the ground
(
y
= 0). Applying Eq. 2-15 to the entire motion (the rise and the fall, of total time
t>
0) we have
y
=
v
0
t
−
1
2
gt
2
=
⇒
t
=
2
v
0
g
which (using our result from part (a)) produces
t
=6
.
4 s. It is possible to obtain this without
using part (a)’s result; one can Fnd the time just for the rise (from ground to highest point) from

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