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**Two-Dimensional Motion Projectiles launched at an angle**

Chapter 3 Two-Dimensional Motion Projectiles launched at an angle

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**Some Variations of Projectile Motion**

An object may be fired horizontally The initial velocity is all in the x-direction vi = vx and vy = 0 All the general rules of projectile motion apply

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**Projectile Motion at an angle**

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**How are they different? Projectiles Launched Horizontally**

The initial vertical velocity is 0. The initial horizontal velocity is the total initial velocity. Projectiles Launched At An Angle Resolve the initial velocity into x and y components. The initial vertical velocity is the y component. The initial horizontal velocity is the x component.

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**Some Details About the Rules**

x-direction ax = 0 vx = vix = vicosΘi = constant x = vixt This is the only equation in the x-direction since there is constant velocity in that direction Initial velocity still equals final velocity

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**More Details About the Rules**

y-direction viy = visinΘi Free fall problem a = g Object slows as it goes up (-9.8m/s2) Uniformly accelerated motion, so the motion equations all hold JUST LIKE STOMP ROCKETS Symmetrical

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**Problem-Solving Strategy**

Resolve the initial velocity into x- and y-components Treat the horizontal and vertical motions independently Make a chart again, showing horizontal and vertical motion Choose to investigate up or down. Follow rules of kinematics equations

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**Solving Launched Projectile Motion**

vi = (r , Θ) = (vix , viy) Horizontal Vertical a = vi = vf = t = x = vix # # = range a = vi = vf = t = x = +/- 9.8m/s2 = g 0 or viy viy or 0 # Max height = y UP or DOWN INVESTIGATION… Where do the resolved components go?

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**Projectile Motion at an angle**

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Example 1: The punter for the Steelers punts the football with a velocity of 27 m/s at an angle of 30. Find the ball’s hang time, maximum height, and distance traveled (range) when it hits the ground. (Assume the ball is kicked from ground level.) Looking for: Total time (t) Max height (y) Range (x) Given: vi = (27m/s, 30o)

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**What do we do with the given info?**

vi = (27m/s, 30o) What are the units? vi = (23.4m/s, 13.5m/s) “resolved” vector m/s 27m/s 30o Viy = 27sin30 Viy = 13.5 Viy = 13.5m/s Vix = 27cos30 Vix = 23.4 Vix = 23.4m/s

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**So where does this info “fit” in the chart?**

Horizontal Vertical a = vi = vf = t = x = 23.4m/s a = vi = vf = t = x = Viy if solving “up” = 13.5m/s Viy if solving “down” = 13.5m/s

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**Pick a “side” to solve – symmetry**

Up: Horizontal Vertical a = vi = vf = t = x = 23.4m/s a = vi = vf = t = y = - 9.8m/s2 13.5m/s vf2 = vi2 + 2gy y = 9.3m vf = vi + gt t = 1.38s

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**Projectile Motion at an angle**

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**Pick a “side” to solve – symmetry**

Down: Horizontal Vertical a = vi = vf = t = x = 23.4m/s a = vi = vf = t = y = 9.8m/s2 13.5m/s vf2 = vi2 + 2gy y = 9.3m vf = vi + gt t = 1.38s

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**Projectile Motion at an angle**

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**On the horizontal side Max height occurs midway through the flight.**

We found t = 1.38s both directions (up and down). How long is the projectile in the air? DOUBLE this time for total air time t = 1.38 x 2 = 2.76s What about range? x = vt x = (23.4)(2.76) x = 64.6m = RANGE

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This tells us… Now we only need an initial velocity vector to determine all of the information we need to have a detailed description of where an object is in its path. vi = (r , Θ) = (vix , viy)

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**Maximum Range vs. Maximum Height**

What angle of a launched projectile gets the maximum height? What angle of a launched projectile gets the maximum range? 90o 45o

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**Projectile Motion at Various Initial Angles**

Complementary values of the initial angle result in the same range The heights will be different The maximum range occurs at a projection angle of 45o

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**Non-Symmetrical Projectile Motion**

Follow the general rules for projectile motion Break the y-direction into parts up and down symmetrical back to initial height and then the rest of the height

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your homework … We are going to see what kind of job Hollywood writers and producers would do on their NECAP assessments… Watching a clip of the Bus Jump, use the timer provided to time the flight of the bus and then do the actual calculations for homework.

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