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Two-Dimensional Motion Projectiles launched at an angle Chapter 3

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Some Variations of Projectile Motion An object may be fired horizontally The initial velocity is all in the x-direction v i = v x and v y = 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 a x = 0 v x = v ix = v i cosΘ i = constant x = v i x t 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 v iy = v i sinΘ i Free fall problem a = g Object slows as it goes up (-9.8m/s 2 ) 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 HorizontalVertical a = v i = v f = t = x = a = v i = v f = t = x = 0 v ix # # = range +/- 9.8m/s 2 = g 0 or v iy v iy or 0 # Max height = y UP or DOWN INVESTIGATION… Where do the resolved components go? v i = (r, Θ) = (v ix, v iy )

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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: v i = (27m/s, 30 o )

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What do we do with the given info? v i = (27m/s, 30 o ) V iy = 27sin30 V iy = 13.5 V ix = 27cos30 V ix = m/s 30 o What are the units? m/s V ix = 23.4m/s V iy = 13.5m/s v i = (23.4m/s, 13.5m/s) “resolved” vector

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So where does this info “fit” in the chart? HorizontalVertical a = v i = v f = t = x = a = v i = v f = t = x = m/s V iy if solving “up” = 13.5m/s V iy if solving “down” = 13.5m/s

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Pick a “side” to solve – symmetry Up: HorizontalVertical a = v i = v f = t = x = a = v i = v f = t = y = m/s - 9.8m/s m/s 0 v f = v i + gt t = 1.38s v f 2 = v i 2 + 2gy y = 9.3m

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Projectile Motion at an angle

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Pick a “side” to solve – symmetry Down: HorizontalVertical a = v i = v f = t = x = a = v i = v f = t = y = m/s 9.8m/s m/s v f = v i + gt t = 1.38s v f 2 = v i 2 + 2gy y = 9.3m

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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. v i = (r, Θ) = (v ix, v iy )

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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? 90 o 45 o

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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 45 o

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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.Bus Jump

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