Projectile Motion - Missouri S&T
Projectile Motion
A projectile is a fired, thrown, or otherwise projected object, such as a bullet, having no capacity for self propulsion (from the American Heritage Dictionary). If it is not self-propelling, then there is no propulsion force acting on the object. There are, however, several other forces—drag, lift, and gravity—that do act on it.
[pic]
Figure ______: Common forces acting Figure ________: Idealized projectile
on a projectile with only gravitational force
How significant are drag and lift forces? Examples of a golf ball and baseball.
Using a computer to compute projectile motion with drag.
Idealized Projectile Motion:
As seen in figure _____, in idealized projectile motion, the only force acting on the particle is gravity, mg, acting downward. If that is the only force, then the only acceleration experienced by the particle is gravity g, acting in the negative y direction.
|Commonly Used Projectile Equations |
|[pic] |
|Coordinate Direction: |x |y |
|Acceleration: |ax = 0 |ay = -g |
| | |where, g = 9.81 m/s2 in S.I. units |
| | |or, g = 32.2 fps2 in U.S. units |
|Velocity: |vx = vox = constant |(2) vy = v0y - gt |
|Position: |(1) x = x0 + vxt |(3) y = y0 + voyt - ½ gt2 |
|An additional equation: | |(4) vy2 = v0y2 - 2g( y - y0 ) |
|Calculating vox and voy, given v0 and |[pic] |(5) vx = vo cos q |
|q0. | |(6) v0y = v0 sin q |
|After launch, calculating vx, vy, |[pic] |vx = vox = vo cos q |
|[pic]and (. | |vy(t) = v0 sin q - gt |
| | |(7) Magnitude: [pic] |
| | |(8) Angle: [pic] |
[pic]
[pic]
|Maximum Range of a Projectile (When y0 = y = 0) | | |
|[pic] |For a given initial velocity, v0, what launch angle,| | |
| |q, will enable the projectile to achieve the maximum| | |
| |range, R? | | |
|Choices of Equations: |Step 1: Find the time that the projectile will land ( y = y0 = 0 ): |
|(1) x = x0 + vxt |Use eqns. (3) and (6): y = y0 + (vosin q)t - ½ gt2 |
|(2) vy = v0y - gt |0 = 0 + (vosin q)t - ½ gt2 |
|(3) y = y0 + voyt - ½ gt2 |Solve for tland: [pic] |
|(4) vy2 = v0y2 - 2g( y - y0 ) |Step 2: Substitute tland into eqn. 1 to get x location at t = tland. |
|where, |x = x0 + vxt |
|(5) vx = vo cos q |x = 0 + (vo cos q)[pic] |
|(6) v0y = v0 sin q |rearranging, [pic], |
| |Using the identity, 2cosqsinq = sin (2q) |
| |we get [pic] |
| |Maximum x will occur when sin 2q is maximized. |
| | |
| | |
|Acceleration: |ax = 0 |ay = -g |
| |(No x forces) |where, g = 9.81 m/s2 in S.I. units |
| | |or, g = 32.2 fps2 in U.S. units |
|Velocity: [pic] |vx = vox = constant |(2) vy = v0y - gt |
|Position: [pic] |(1) x = x0 + vxt |(3) y = y0 + voyt - ½ gt2 |
|An additional equation: |(Not applicable) |(4) vy2 = v0y2 - 2g( y - y0 ) |
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