Projectile Motion NotesProjectile Motion Notes

Projectile MotionProjectile Motion

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

Projectile MotionProjectile Motion

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

Projectile MotionProjectile Motion When an object is thrown or launched When an object is thrown or launched

into the air, it undergoes projectile into the air, it undergoes projectile motion. The pathway of any projectile motion. The pathway of any projectile is a parabola (excluding air resistance).is a parabola (excluding air resistance).

Projectile MotionProjectile Motion After breaking the initial velocity into x After breaking the initial velocity into x

and y components separately, the and y components separately, the pathway of the object can be tracked by pathway of the object can be tracked by using a series of equations:using a series of equations:

Projectile Motion EquationsProjectile Motion Equations

Variables:Variables: Y = distance in meters verticallyY = distance in meters vertically VVyy = y component of velocity = y component of velocity

VVoyoy = y component of initial velocity = y component of initial velocity

VVoxox = x component of initial velocity = x component of initial velocity g = acceleration of gravityg = acceleration of gravity t = time during flight t = time during flight

Projectile Motion EquationsProjectile Motion Equations

Vertical:Vertical:

Velocity: Velocity: vvyy = v = vo(y)o(y) – g t – g t

Distance: Y = Distance: Y = vvo(y)o(y) t – ½ g t t – ½ g t22

Time for maximum height: Time for maximum height: tt = = vvo(y)o(y) / g / g

Projectile Motion EquationsProjectile Motion Equations

Horizontal:Horizontal: Velocity: VVelocity: Vx x remains constant;remains constant;

therefore therefore vvo(x)o(x) = = vvf (x)f (x)

Distance: Distance: X =X = vvo(x)o(x) t t

Projectile MotionProjectile Motion

In projectile problems, vertical and horizontal In projectile problems, vertical and horizontal motions are completely independent of one motions are completely independent of one another. another.

For example: An object will accelerate For example: An object will accelerate downward at the exact same rate regardless of downward at the exact same rate regardless of whether it is dropped, thrown forward slowly, or whether it is dropped, thrown forward slowly, or thrown forward very quickly. thrown forward very quickly.

Likewise: An object will continue to move Likewise: An object will continue to move forward the same distance every second whether forward the same distance every second whether it is rolling along a flat surface or moving it is rolling along a flat surface or moving vertically.vertically.

Projectile MotionProjectile Motion

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

Projectile Motion EquationsProjectile Motion Equations

One method for solving for horizontally One method for solving for horizontally projected objects: projected objects:

1) Write down all variables, and identify 1) Write down all variables, and identify them as horizontal or vertical.them as horizontal or vertical.

2) Solve one dimension (x or y, depending 2) Solve one dimension (x or y, depending on what is given in the problem) for time.on what is given in the problem) for time.

3) Use that time in the other dimension to 3) Use that time in the other dimension to find unknown quantityfind unknown quantity

Practice ProblemsPractice Problems

Turn to page 129, work through problem on Turn to page 129, work through problem on right side of pageright side of page

Asking for vertical distance (y) and horizontal Asking for vertical distance (y) and horizontal distance (x)distance (x)

Initial vertical velocity = 0 m/s, Initial Initial vertical velocity = 0 m/s, Initial horizontal velocity = 20 m/s, t = 2 seconds, horizontal velocity = 20 m/s, t = 2 seconds, g = -9.81 m/sg = -9.81 m/s22

Solve for y, solve for xSolve for y, solve for x