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)

 

 

PLEASE ANSWER:

            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?

TODAY:

            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.

 

OTHER NOTES:

Strobe diagrams:

Path of motion:
PROJECTILE FORMULAS:

 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.