Two Dimensional Motion, with gravity…..


NOTES, fill in the blanks as we go over in class (Tue, Nov 29, 2005)

Due Thur Dec 1, Read Section 3-3, do Practice 3D pg 102, Sec Review pg 105 1-5. (plus 6,7 for Honors).

Due Fri Dec 2. Projectile 2 Practice Problems

Due Fri Dec 2, Projectile Lab1 Procedures (see pg 120,121 for ideas)

Due Mon Dec 5 Review questions Chapter 3

Tue Dec 6th Test (Relative Motion, Relativity, Vectors, Projectiles)

Thur Dec 8th, Projectile Labpt1 due (indiv)




            Why do we use vectors at right angles to represent projectile motion?

            If I roll something off the table (without air resistance), what force is acting on it? In what direction?

What will its final horizontal velocity be?


            Independence of horizontal and vertical motion





We separate motion in two dimensions into x and y components…. Horizontal (x) usually has no force, thus no acceleration. Vertical (y) has the acceleration due to gravity.


Projectile motion: 2 dimensions, only force is that of gravity (no air resistance), makes a parabola..


What property is the same between the two???




Steps for solving projectile problems:


1) Draw a picture!!      2) Write down the info given as x and y parts. (Change vector velocity to x and y components if needed)        3) Determine what needs to be solved. 4) Find the time in one direction and use it in the other direction.         5) Solve.



Strobe diagrams:

Path of motion:

 Pythagorean:        V2= Vx2 + Vy2                      

                                                                        Vx= V cos (Ø)                                  

                                                                         Vy= Vsin (Ø)                                   

                                                                        Ø= tan-1(Vy/Vx)



            X                                                                     Y                    


Ax = 0 so                                                                    Ay= -9.8 m/s2

Vix=Vfx=Vavgx                                                        Viy= Vsin (Ø)


Dx = Vx T                                                                  h=Dy=ViyT + ½ Ay T2

                                                                                    Vfy =Viy + Ay T

Vx= V cos (Ø)                                                                        Vfy2=Viy2 + 2 Ay Dy



If launched horizontally, Viy =0.


If launched at an angle so it hits the ground, at top Vy=0, at ground Dy = 0.