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point shown (1.25 s after the ball is released) which is when the ball returns to its original
height. In English units,
g
= 32 ft/s
2
.
(a) Using
x
–
x
0
=
v
x
t
we obtain
v
x
= (40 ft)/(1.25 s) = 32 ft/s. And
yy
v
t g
t
y
−=
=
−
00
1
2
2
0
yields
()
2
1
0
2
32 ft/s
1.25 s
20 ft/s.
y
v
==
Thus, the initial speed is
22


(32 ft/s)
(20 ft/s)
38 ft/s.
vv
+
=
G
(b) Since
v
y
= 0 at the maximum height and the horizontal velocity stays constant, then
the speed at the top is the same as
v
x
= 32 ft/s.
(c) We can infer from the figure (or compute from
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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

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