Introduction to 2D Projectile Motion

Projectile Motion An example of 2-dimensional motion. Something is fired, thrown, shot, or hurled near the earth’s surface. Horizontal velocity is constant. Vertical velocity is accelerated. Air resistance is ignored.

Trajectory of Projectile x y This projectile is launched at an angle and rises to a peak before falling back down.

Trajectory of Projectile x y The trajectory of such a projectile is defined by a parabola.

Trajectory of Projectile x y Range The RANGE of the projectile is how far it travels horizontally.

Trajectory of Projectile x y Maximum Height Range The MAXIMUM HEIGHT of the projectile occurs halfway through its range.

Trajectory of Projectile x y g g g g g Acceleration points down at 9.8 m/s2 for the entire trajectory.

Position graphs for 2-D projectiles x y t

To work projectile problems… …you must first resolve the initial velocity into components. Vo,y = Vo sin  Vo  Vo,x = Vo cos 

Trajectory of Projectile x y v v v vo vf Velocity is tangent to the path for the entire trajectory.

Trajectory of Projectile x y vx vy vx vy vx vy vx vx vy The velocity can be resolved into components all along its path.

Trajectory of Projectile x y vx vy vx vy vx vy vx vx vy Notice how the vertical velocity changes while the horizontal velocity remains constant.

Trajectory of Projectile x y vx vy vx vy vx vy vx vx vy Where is there no vertical velocity?

Trajectory of Projectile x y vx vy vx vy vx vy vx vx vy Where is the total velocity maximum?

2D Motion Resolve vector into components. Position, velocity or acceleration Work as two one-dimensional problems. Each dimension can obey different equations of motion.

Horizontal Component of Velocity Newton's 1st Law Is constant Not accelerated Not influence by gravity Follows equation: x = Vo,xt

Horizontal Component of Velocity

Vertical Component of Velocity Newton's 2nd Law Undergoes accelerated motion Accelerated by gravity (9.8 m/s2 down) Vy = Vo,y - gt y = yo + Vo,yt - 1/2gt2 Vy2 = Vo,y2 - 2g(y – yo)

Horizontal and Vertical

Horizontal and Vertical

Launch angle vo Zero launch angle

Launch angle  vo Positive launch angle

Symmetry in Projectile Motion  vo - vo Negligible air resistance Projectile fired over level ground Launch and Landing Velocity

Symmetry in Projectile Motion to = 0 Time of flight

Symmetry in Projectile Motion to = 0 2t Projectile fired over level ground Time of flight Negligible air resistance

Problem Solution Strategy: 1. Upward direction is positive. Acceleration due to gravity (g) is downward thus g = - 9.8 m/s2 2. Resolve the initial velocity vo into its x and y components: vox = vo cos θ voy = vo sin θ

Problem Solution Strategy (pt 2) 3. The horizontal and vertical components of its position at any instant is given by: x = voxt y = voy t +½gt2 4. The horizontal and vertical components of its velocity at any instant are given by: vx = vox vy = voy + gt

Problem Solution Strategy (pt 3) 5. The final position and velocity can then be obtained from their components.