This preview shows page 1. Sign up to view the full content.
100. (a) At the highest point, the velocity
v
=
v
x
is purely horizontal and is equal to the
horizontal component of the launch velocity (see section 46):
v
o
x
=
v
o
cos
θ
, where
30
=°
in this problem. Eq. 817 relates the kinetic energy at the highest point to the
launch kinetic energy:
K
o
=
mg y
+
1
2
mv
2
=
1
2
mv
o
x
2
+
1
2
mv
o
y
2
.
with
y =
1.83 m. Since the
mv
ox
2
/2 term on the lefthand side cancels the
m
v
2
/2 term on
the righthand side, this yields
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, Energy, Kinetic Energy

Click to edit the document details