This preview shows page 1. Sign up to view the full content.
Unformatted text preview: etric equations
cos
1
2
We may eliminate the parameter by solving the first equation for :
sin cos
Thus,
sin cos 1
2 cos Simplifying, This is a parabola. Therefore, the path the projectile follows is a parabolic one. Early in
the history of the study of projectile motion if was believed that projectiles followed a path 9 composed of an initial rectilinear motion, followed by a second stage circular motion, and
finally another rectilinear motion (see figure below).
Circular This is what we believed projectile trajectories looked like
Example 7 A bomber traveling at 640 mph releases a missile from 35,000 feet. The missile
malfunctions upon launch and falls freely under the force of gravity. The intended target lies 10
miles ahead at the time of release. Will the missile hit its target?
Solution 640 35,000 ft. Our objective is to determine if
10 miles. Since we will work in English units, we need
to express distances in feet and speeds in ft/sec. 10 miles = 52,800 feet. Also, 640 mph =
938.67 ft/sec is the initial speed of the missile which is inherited from the bomber.
The position vector for the missile is 1
2
938.67,0 . We have
0 Where 0 0,35000 , 0 0 16
or
938.67
35,000 16 In order to find
we need to know . When the missile strikes the ground,
Therefore, the time of flight is 0. 35,000
46.77 sec
16
The horizontal distance traveled during this time is
938.67 46.77 43,901.60 feet This is short by a several thousand feet of the desired 52,800 feet. The missile will miss its
mark. 10 Unit Tangent, Unit Normal, and Binormal Vectors
The study of more general forms of motions is greatly facilitated by the introduction of three very
important vectors: the Unit Tangent, Unit Normal, and Binormal vectors. These vectors determine
the intrinsic properties of curves in general and in particular provide a better framework for the
theoretical study of motion.
The Unit Tangent Vector
Let a curve C be represented by the vector function
. We learned that at points where
defined by
is tangent to the curve. Therefore, the vector
the vector
unit tangent
. is a unit vector. We call it the unit tangent vector to the curve at
, Example 8 Find the unit tangent vector of 0. at Solution
1,2
At √1 0 1
4 1 0 4
2 √1
1 4 √1
, , 2
√1 4 2
√1 4 ,
√5 √5 Verify that thi...
View
Full
Document
This note was uploaded on 10/24/2012 for the course MAC 2313 taught by Professor Lopez during the Spring '10 term at Miami Dade College, Miami.
 Spring '10
 LOPEZ
 Equations

Click to edit the document details