Projectiles AP Physics C Mrs. Coyle

Horizontal Projectiles are launched with a horizontal initial velocity. http://www.physicsclassroom.com/mmedia/vectors/hlp.cfm

Motion in Two Dimensions http://www.physicsclassroom.com/mmedia/vectors/hlp.cfm

A horizontal projectile from the same height Which falls first? A free falling object Or A horizontal projectile from the same height

A horizontal projectile Which falls first? A horizontal projectile Or A horizontal projectile from the same height with a higher initial horizontal speed

Question A zoologist is aiming at a monkey with a tranquilizer gun. Both are at the same level. The monkey is about to jump. Should the zoologist aim horizontally or low in order to tranquilize the monkey?

Horizontal Projectiles Two dimensional motion Initial velocity,vx ,is entirely horizontal X-axis : constant velocity Y-axis : constant acceleration

A Look at the Velocities Vx= constant Vyi=0 Vy = a t

Characteristics of Path Trajectory Parabola Range

Equations for Horizontal Projectiles x=vx t ____ y= 1 at2  t= √2y/a 2 vY=at

Projectiles at an Angle

Projectiles at an Angle http://www.physicsclassroom.com

Projectiles at an Angle Motion in two dimensions x-axis constant velocity y-axis constant acceleration Parabolic trajectory

Initial Velocity vxi=vcosq vyi=vsinq v

A Look at the Velocities

Equations for Projectiles at an Angle Initial Velocity is at an angle with the horizontal vx=constant vy=vyi+at a or g=-9.8m/s2 Dy=vyit+ 1at2 2

Complementary angles have the same range.

Additional Key Points At the max height: vy=0 The entire trip vx=constant=vxi Maximum range when q=45o in the absence of air resistance. Complementary angles result in the same range. Symmetry Resultant velocity is tangent to the curve at the point of study. Ignore air resistance.

Prove the maximum height equation (for full symmetric motion)

Prove the range equation Hint: Use the identity sin2θ=2sinθcosθ Why does max range occur at 45o?

Non symmetric projectile Ex: Find t and vf. Ans: 4.22s, 35.9m/s