This preview shows page 1. Sign up to view the full content.
2
1
00
2
0.
f
yf
pf
y
yv
t
g
t
−=
=
−
This gives us
v
0
y
= 1.25
g
p
, and we see we need another
equation (by analyzing another point, say, the nexttolast one)
2
1
2
yp
y
t
g
t
−
with
y
= 6 and
t
= 2; this produces our second equation
v
0
y
= 2 +
g
p
. Simultaneous
solution of these two equations produces results for
v
0
y
and
g
p
(relevant to part (b)). Thus,
our complete answer for the initial velocity is
ˆˆ
(10 m/s)i
(10 m/s)j .
v
=+
G
(b) As a byproduct of the part (a) computations, we have
g
p
= 8.0 m/s
2
.
(c) Solving for
t
g
(the time to reach the ground) in
yy
v
t
g
t
g
y
gp
g
== +
−
0
1
2
2
leads to a
positive answer:
t
g
= 2.7 s.
(d) With
g
= 9.8 m/s
2
, the method employed in part (c) would produce the quadratic
equation
2
4.9
10
2
0
gg
tt
−+
+
=
and then the positive result
t
g
= 2.2 s.
This is the end of the preview. Sign up
to
access the rest of the document.
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, Acceleration

Click to edit the document details