47. We neglect air resistance for the duration of the motion (between “launching” and
“landing”), so
a
= –
g
= –9.8 m/s
2
(we take downward to be the –
y
direction). We use the
equations in Table 21 (with
Δ
y
replacing
Δ
x
) because this is
a
= constant motion.
(a) At the highest point the velocity of the ball vanishes. Taking
y
0
= 0, we set
v
= 0 in
vv
g
y
2
0
2
2
=−
, and solve for the initial velocity:
vg
y
0
2
=
.
Since
y
= 50 m we find
v
0
= 31 m/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. 215 to the entire motion (the rise and the fall, of total time
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler
 Physics, Resistance

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