Agenda 1) Warm-Up 5 min 2) Turn-in Blue “Vector” Folder! 3) Vocab. Words 10 min 4) Finish Projectile Motion fill-in- blank Notes. 10 min 5) New Formulas 5 min 6) Example Problems 5 min 7) Virtual Lab: Projectiles at an angle 45 min 8) Take-Home Quiz m/s

Vocab. 1.Axis: the straight line about which rotation takes place. 2.Rotation: also called spin; the spinning motion that takes place when an object rotates about an axis located within the object. 3.Revolution: motion of an object turning around an axis outside the object.

Projectile fired at an Angle: When a projectile is fired at an angle with the horizontal, the principle of independent velocities still holds. The initial velocity of the projectiles can be “resolved” into 2 components. One component is directed vertically and the second is directed horizontally. These components are treated separately when solving problems.

Horizontally: Look at all the horizontal vectors---They are all equally spaced apart. There is still no acceleration in the horizontal direction so the cannonball moves equal horizontal distances in equal time intervals---constant velocity.

Vertically: Look at all the vertical vectors---They get shorter and shorten then disappear at the top of the path, then get longer and longer. This is because there is acceleration vertically (in the direction of the earth’s gravity). The vertical velocity and therefore distance gets bigger and bigger each second the object is still moving. Notice that the horizontal component vector is the same length at any point along the cannonball’s path. The vertical component gets smaller then, disappears then, gets bigger.

These vectors represent the horizontal and vertical component of velocity. The ACTUAL velocity of the object is represented by the diagonal of the parallelogram formed by the components. At the top of the path, the vertical component vanishes, or becomes zero, so the ACTUAL velocity of the cannonball at the top of the path is the exact same as the horizontal velocity at all other points (instead of a combination of the horizontal and vertical).

The above picture has the same launching speed as the last picture and is the same idea as the last picture except the object is projected at a steeper angle. Notice the initial velocity vector has a greater vertical component than in the last picture when the projection angle was smaller. This bigger vertical component results in a higher path, but the horizontal component is less so the range is less.

DUE TODAY: Projectile fill-in-blanks (KEEP) Virtual Lab Blue “Vector” Unit Folder DUE NEXT CLASS: Projectile Motion [Take-Home Quiz] Study for Vocab. Quiz Bring Calculator & Protractor! “There is no better high than discovery.” E. O. Wilson

 Projectile motion under the influence of near-Earth--- gravity will produce a curved path.  The vertical direction is on the “y “axis. The only force acting on any projectile vertically will be gravity. The acceleration will be the constant g = m/s 2.

 KEY: What happens to a particle vertically does not affect its horizontal motion. The opposite is also true. The axes are independent.  NOTE what axis is CHANGING!!

Those projectiles that accelerate only in a vertical direction while moving at a constant horizontal velocity have a path that forms a PARABOLA.

 The range is the horizontal distance.  Solve for the time needed for the projectile to reach its maximum height (remember that gravity is acting) by using the “vertical” distance.  Once you know the time of flight, solve for the range by using: D H = Horizontal velocity X Time

 Satellite – an object that falls around Earth or some other body rather than falling into it because of its tremendous speed. ◦ Page 127 (Hewitt book)

TWO DIMENSIONAL MOTION  If no force acts horizontally, the motion found is at constant velocity- due to inertia.  The equation of motion is then D H = V H t.  BUT, to find V H when you have 2- dimensional motion, resolve the initial velocity into its horizontal component by using the formula : V H = V i cos 

 To find V v, resolve the initial velocity into its vertical component.  The formula to do this is: V v = V i sin Ø

 The time an object moves horizontally is the same as the time the object moves vertically.  So, Total time is the “time in the air” for a projectile at-an-angle.

 To find time: Remember that gravity is affecting the motion and therefore, the amount of time the projectile is in the air for “at an angle” will be TOTAL time.  So, t total = V v / g so D H = V H ( t t )

A vertical and a horizontal component. With the trajectory being the initial velocity*

 Acceleration is constant for a projectile.  Speed and velocity change at each point along the parabolic pathway.  What is the speed of the projectile at the very top of its pathway?  What is the acceleration of the projectile at the very top of its pathway?