This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ctor and the time it takes for the projectile to hit
the ground. What is the highest point in the trajectory?
Solution 1
2
We just need to figure out what we are given. Also, since we assume the motion takes place
in a plane, we may assume it is the xy plane and write
0 0 1
2
since the projectile is fired from ground level and
0 with 0 0 7 0 400 cos 45° , 400 sin 45° 282.84 Thus,
282.84
282.84 1
2 282.84
282.84 282.84 1
2 The parametric equations of the trajectory are
282.84
1
2
0. That means 282.84
The projectile hits the ground when
1
2
We may solve this equation by factoring :
282.84 0 1
2 0 16 282.84 0 or
282.84 Thus,
0 sec or
17.68 sec The first value corresponds to when the object is fired
from ground level. The second value is, of course, the time it takes for the projectile to hit
the ground again.
How high does the object reach? When the object reaches its highest point, its vertical
component of velocity is 0. This vertical component is given by
1
2 282.84
282.84
At its highest point 0. Therefore,
282.84 Solving for : 0 282.84 8.84 ft Notice that this is the same as half the time for the full trajectory: 17.68/2
this always be the case? Think about it carefully. 8.84. Will ft/sec at an angle of
Example 6 Suppose a projectile is fired from ground level with a speed of
with the horizontal. Determine (1) the range of the projectile, that is, the horizontal distance it
travels before striking the ground. (2) for what angle will this range be a maximum? (3) What is
the shape of the projectile’s trajectory?
Solution
Again, we may assume a trajectory that is confined to the xy plane, in which case the
general solution to the problem is
1
0
0
2 8 which we may write in component form:
cos
1
2
Placing the origin of our coordinate system at the point where the projectile is fired,
0. Therefore,
cos
sin 1
2
In order to determine the range, we need to find the value of when the projectile hits the
ground. This happens when
0. Thus, the time of flight is found from the equation
sin sin 2 sin 0 or
The horizontal distance traveled in is:
cos Using the identity sin 2 0 16 Solving for : · 2 sin 2 sin cos , This is the range of the projectile. (2) What is its maximum value? It is and it occurs when sin 2 1, that is, when 2 /2 or when /. Finally, (3) what path does the projectile follow? consider the param...
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