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#### Presentation Transcript

**1. **Projectile Motion Notes

**2. **Projectile Motion A vector can be broken down into component vectors by creating a right triangle. This is done to evaluate motion in two dimensions.

**3. **Projectile Motion In order to determine the magnitude and direction of the component vectors, Trigonometry and the Pythagorean Theorem must be used.

**4. **Projectile Motion When an object is thrown or launched into the air, it undergoes projectile motion. The pathway of any projectile is a parabola (excluding air resistance).

**5. **Projectile Motion After breaking the initial velocity into x and y components separately, the pathway of the object can be tracked by using a series of equations:

**6. **Projectile Motion Equations Variables:
Y = distance in meters vertically
Vy = y component of velocity
Voy = y component of initial velocity
Vox = x component of initial velocity
g = acceleration of gravity
t = time during flight

**7. **Projectile Motion Equations Vertical:
Velocity: vy = vo(y) – g t
Distance: Y = vo(y) t – ½ g t2
Time for maximum height: t = vo(y) / g

**8. **Projectile Motion Equations Horizontal:
Velocity: Vx remains constant;
therefore vo(x) = vf (x)
Distance: X = vo(x) t

**9. **Projectile Motion In projectile problems, vertical and horizontal motions are completely independent of one another.
For example: An object will accelerate downward at the exact same rate regardless of whether it is dropped, thrown forward slowly, or thrown forward very quickly.
Likewise: An object will continue to move forward the same distance every second whether it is rolling along a flat surface or moving vertically.

**10. **Projectile Motion Time is the only variable which will be the same for the two motions. The time it takes for an object to hit the ground is the same amount of time it has to move forward.

**11. **Projectile Motion Equations One method for solving for horizontally projected objects:
1) Write down all variables, and identify them as horizontal or vertical.
2) Solve one dimension (x or y, depending on what is given in the problem) for time.
3) Use that time in the other dimension to find unknown quantity

**12. **Practice Problems Turn to page 129, work through problem on right side of page
Asking for vertical distance (y) and horizontal distance (x)
Initial vertical velocity = 0 m/s, Initial horizontal velocity = 20 m/s, t = 2 seconds, g = -9.81 m/s2
Solve for y, solve for x