# Unit 4 Class Notes Accelerated Physics Projectile Motion Days 1 thru 3.

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Unit 4 Class Notes Accelerated Physics Projectile Motion Days 1 thru 3

Day #1 Free-Falling Object Review

In-Class Practice

Day #2 Horizontal Projectiles

Horizontal Projectiles The first example on the previous slide is often called a horizontal projectile. This is the first example we will look at. – The object is given an initial horizontal velocity. Thus, velocity is purely horizontal (no y component) – Our initial velocity in the y direction v iy =0. – Next, analyze the motion in the horizontal and vertical direction separately.

Figure #1 Figure #2 Figure #3 From Reading Assignment handout

Using figures from previous slide Figure 1 – Free fall object. The object’s position after every second of falling is shown. The object is accelerating because it travels a longer distance in each successive time interval. Figure 2 – Horizontally traveling object at a constant velocity. Acceleration equals zero so each time interval shows the same distance traveled. Figure 3 – A combination of both these motions. When a projectile falls, it accelerates downward so its y direction matches figure 1. However, since nothing (including air) slows the object down as it moves horizontally, its x direction matches figure 2.

Since projectile motion is a combination… We will analyze the motion separately. The only shared variable between the two motions is TIME (since they both occur in the same amount of time) Solve for time in vertical (y) side, then plug in to horizontal side. Horizontal (x)Vertical (y) a x =0 v ix =v fx = constant = v x ∆x = v x ∆t (distance = velocity x time) a y =-9.8 m/s 2 ∆y = 1 / 2 a y (∆t) 2 +v iy ∆t V fy 2 = v iy 2 + 2a y ∆y A = (v fy – v iy ) /∆t ∆y = ½∆t(v iy +v fy )

Horizontal (x) Vertical (y) From Reading Assignment handout There are so few equations in the x direction because a=0 so velocity is constant. In the y direction we have the 4 familiar acceleration formulas. EXAMPLE 1: A ball is rolled off a flat roof hat is 30 m above the ground. If the ball’s initial speed is 15 m/s, find the time needed to strike the ground as well as the distance that it lands away from the building.

Horizontal (x) Vertical (y) From Reading Assignment handout

Horizontal (x) Vertical (y) From Reading Assignment handout

In-Class Practice

Day #3 COMPETITION LAB!!!! (Ball Rolling Off Table)

Day #4 Tracking and Impact Velocity

Side-ways Toss (Gravity Turned Off) Each position corresponds to 1 sec later than the previous position (starting with the red dot)

Horizontal Projectiles Side-ways Toss (Gravity Turned Off) Free-Fall

Horizontal Projectile Side-ways Toss (Gravity Turned Off) Free-Fall Velocities Impact Velocity

 V fx V fy V impact

Horizontal (x) Vertical (y) 19.6 m/s 4 m/s V2V2  V 2 = 20 m/s [78.46 o BH]

In-Class Practice

Day #5 Angled Projectiles on Flat Ground

What we’ve done so far What we still need to do

Projectiles at Angles Side-ways Toss (Gravity Turned Off) Throw-up, then Free-Fall

So, what are the Equations we use for this “NEW” type of projectile???? Vertical Horizontal There are NO NEW EQUATIONS

 vivi  vivi v ix v iy = v i sin  = v i cos  So…..How are “angled” projectiles different than “horizontal” projectiles? V iy = 0 (for horizontal projectiles) V iy = something, + or - (for horizontal projectiles)

 vivi What are some important things to remember? The velocity at every moment in time is the resultant of the two velocity components V ix = V fx = V x = 0 (since a x = 0) When rising, V y is positive (when falling = V y is negative) V y, top = 0 If launching and landing occurs at the same height (on level ground) then the launching/landing speeds & angles are the same 

In-Class Practice

Day #6 MORE Angled Projectiles … Sometimes NOT on Flat Ground

In-Class Practice

Day #7 Field Goal Style Problems

In-Class Practice

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