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 6^{th} Test (Relative Motion, Relativity,
Vectors, Projectiles)

Thur Dec 8^{th}, 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: **V**^{2}= Vx^{2} + Vy^{2}

Vx=
**V** cos (Ø)

Vy= **V**sin (Ø)

Ø=
tan-1(Vy/Vx)

__X
__ __Y __

Ax = 0 so Ay=
-9.8 m/s^{2}

Vix=Vfx=Vavgx Viy=** V**sin (Ø)

Dx = Vx T h=Dy=ViyT
+ ½ Ay T^{2}

Vfy
=Viy + Ay T

Vx= **V** cos (Ø) Vfy^{2}=Viy^{2}
+ 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.