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**Chapter 5 Projectile motion**

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**Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion**

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**Chapter 5: considers motion that follows a diagonal path or a curved path**

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**When you throw a baseball, the trajectory is a curved path.**

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**We are going to separate the motion of a projectile into independent x and y motions**

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**The vertical motion is not affected by the horizontal motion**

The vertical motion is not affected by the horizontal motion. And the horizontal motion is not affected by the vertical motion.

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Observe: a large ball bearing is dropped at the same time as a second ball bearing is fired horizontally.

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What happened?

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**Remember adding 2 perpendicular vectors horizontal and vertical vectors.**

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**When we add perpendicular vectors we use Pythagorean theorem to find the resultant.**

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**Consider a vector B that is pointed at an angle q wrt horizontal direction.**

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**We are going to break vector B into 2 perpendicular vectors: Bx and By**

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**if you ADD vectors Bx + By you get vector B.**

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**Graphically, we can say:**

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**Draw a rectangle with vector B as the diagonal**

Draw a rectangle with vector B as the diagonal. the component vectors Bx and By are the sides of the rectangle

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Application:

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A Boat in a river

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**How can we describe the motion of a boat in a river?**

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**The motion is affected by the motor of the boat and by the current of the river**

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**Imagine a river 120 meters wide with a current of 8 m/sec.**

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**Imagine a river 120 meters wide with a current of 8 m/sec.**

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**If a boat is placed in the river [motor is off] , how fast will the boat drift downstream?**

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**If the boat is drifting, the total speed of the boat just equals the speed of the current.**

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**for a boat drifting with the current: Vtotal = Vboat + Vcurrent Vtotal = -0 + Vcurrent = 8 m/sec**

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**Now suppose this boat can travel at a constant 15 m/sec when the motor is on .**

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**What is the total speed of the boat downstream when the motor is on?**

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**The boat is traveling in the same direction as the current.**

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**V total downstream = Vtotal = Vboat + Vcurrent Vtotal = 15↓ + 8↓ = 23 m/sec ↓**

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**What is the total speed of the boat traveling upstream [against the current] ?**

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**The boat and current now move in opposite directions**

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**Vtotal = Vboat +Vcurrent Vtotal = 15 ↓+ ( 8 ↑) Vtotal = 7↓ m/sec**

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**Summary: traveling downstream: Vboat + Vcurrent Traveling upstream: Vboat + [-Vcurrent]**

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Crossing the river.

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**If there was no current, how many seconds needed for this boat to travel 120 meters from A to B?**

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**Velocity = distance time so time = distance velocity**

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**time = distance velocity time = 120 m 15 m/sec time = 8 seconds**

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**But there if IS a current**

But there if IS a current. what happens when you try to go straight across the river from A to B?

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**The boat will travel from A to C.**

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Every second the boat travels ACROSS 15 meters and AT THE SAME TIME every second the boat will be pushed DOWNSTREAm 8 meters by the current .

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**Vboat = 15 and Vcurrent = 8↓ These velocities are perpendicular**

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**The RESULTANT velocity of the boat is**

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Vresultant2 = Vboat2 + Vcurrent2 Vresultant2 = Vresultant2 = =289 Vresultant = 17 m/sec

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**The boat still crosses the river in 8 seconds , but it lands downstream at point C not at point B.**

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**How far downstream is point C?**

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**Since the boat travels for 8 seconds the current pushes the boat for 8 seconds Vcurrent = 8 m/sec**

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Velocity = distance/time so Distance = Velocity• time Distance = 8 m/sec • 8 sec distance downstream = 64 meters

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**What is the total distance the boat travels? D2 = Dx2 + Dy2**

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D2 = D2 = D2 = D = 136 meters

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**The triangles are similar:**

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REMEMBER Every second the boat travels 15 meter across in the x direction IT ALSO TRAVELS 8 meter in the y direction

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**What if you want to travel from point A to point B? Can you do that?**

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**You can cross from A to B if you point the boat in the correct direction.**

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Remember: Two perpendicular vectors can be added to produce a single resultant vector that is pointed in a specific direction.

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SIMILARLY ANY vector at angle q can be broken into the sum of two perpendicular vectors: one vector only in x direction and one vector only in y direction.

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**The magnitude of the component vectors is given by Vx = Vocosq Vy = Vosinq**

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If you want to travel from A to B, you must direct the boat so that the “y component” of the boat’s velocity cancels the velocity of the current.

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**Point the boat so that the component of the boat’s velocity “cancels” the river**

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**Choose VboatY so that it is equal and opposite to the Vcurrent**

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**VboatX = Vboatcosq VboatY = Vboatsinq**

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**How do we find angle q, the direction to point the boat?**

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Use arcsin or arctan

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**arcsin = sin-1 Arcsine means “ the angle whose sine is” : Sin-1 [VboatY/Vboat] = q**

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**Arctan = Tan-1 Arctan means “ the angle whose tangent is”**

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**Arctan = Tan-1 Arctan means “ the angle whose tangent is” tan-1[Vboaty/Vboatx] = q**

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Remember

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PROJECTILE MOTION

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Projectile motion: A projectile that has horizontal motion has a parabolic trajectory We can separate the trajectory into x motion and y motion.

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**In the x direction: constant velocity Vx = constant distance in x direction X = Vx • t**

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**In y direction: free fall = constant acceleration**

In y direction: free fall = constant acceleration. Velocity in y direction : V = Vo – g t Distance in y direction Y = Yo + Vot – ½ g t2

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**The range of a projectile is the maximum horizontal distance.**

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**Range and maximum height depend on the initial elevation angle.**

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**If you throw a projectile straight up, the range = 0 height is maximum**

If you throw a projectile straight up, the range = 0 height is maximum. 0 degrees : the minimum range but the maximum height.

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**The maximum range occurs at elevation 45o**

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**And for complementary angles 40 and 50 degrees 30 and 60 degrees 15 and 75 degrees 10 and 80 degrees**

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The range is identical for complementary angles BUT the larger elevation angle gives a greater maximum height.

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Remember:

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For a horizontal launch: Vo = initial horizontal velocity 0 = initial vertical velocity in x direction: velocity is constant in y direction: acceleration is constant

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If one object is fired horizontally at the same time as a second object is dropped from the same height, which one hits the ground first?

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Horizontal launch: in x [horizontal] direction velocity is constant Vx = Vo acceleration = 0 range = Vo • t [t = total time ]

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**Horizontal launch: In y direction: projectile is free falling**

Horizontal launch: In y direction: projectile is free falling. Voy = 0 Acceleration = g = 10 m/sec2↓ V = gt ↓ d = ½ gt2

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**Projectile motion lab: part 1**

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**Part 1: determine the velocity Vo of the projectile.**

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**Projectile: - fired horizontally from height h**

Projectile: - fired horizontally from height h. - follows parabolic path - Range R is where projectile hits the floor.

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**Equations: in y direction Voy = 0 g = constant acceleration distance h = ½ gt2**

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**Equations in x-direction acceleration = 0 [constant velocity] V= Vo Range R = Vo ▪ t**

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**Part 1: fire projectile horizontally. Measure all distances in METERS**

Part 1: fire projectile horizontally. Measure all distances in METERS. Measure starting height , h. Measure range R.

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Part I Calculations:

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**distance h = ½ gt2 measure h [ in METERS] use g = 10 m/sec2**

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**solve equation to find t [ in seconds ] distance h = ½ gt2 h= 5 t 2**

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**Measure value for R, the range in x direction in METERS**

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**Use equation: R = Vo ∙ t use measured value of R and calculated value for T**

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**Example: A projectile is fired horizontally from a table that is 2**

Example: A projectile is fired horizontally from a table that is 2.0 meters tall. The projectile strikes the ground 3.6 meters from the edge of the table.

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**Given: H = 2.0 meters R = 3.6 meters**

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h = ½ g t2 R = Vo t

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h = ½ g t = ½ [10] t2

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2.0 = ½ [10] t2 2.0 = 5 t2 2/5 = 0.40 = t2

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t = 0.63 sec

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**For equation: R = Vo t use R = 3.6 m and t = 0.63 sec**

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**R = 3.6 m = Vo [.63 sec] Vo = 3.6 m 0.63 sec Vo = 5.7 m/sec**

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**The height of a projectile at any time along the path can be calculated.**

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**First calculate the height if there was no gravity**

First calculate the height if there was no gravity. If that case, a projectile would follow a straight line path

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**the projectile is always a distance 5t2 below this line.**

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Y = voy t – ½ gt2 Y = voy t – 5t2 i

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summary

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**Vectors have magnitude and direction Scalars have only magnitude**

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The resultant of 2 perpendicular vectors is the diagonal of a rectangle that has the 2 vectors as the sides.

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**The perpendicular components of a vector are independent of each other.**

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The motion of a boat in a stream is the sum of a constant velocity of a boat [x dir] and the constant velocity of the stream [y dir]

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**The path of a boat crossing a stream is diagonal**

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The horizontal component of a projectile is constant, like a ball rolling on a surface with zero friction. Objects in motion remain in motion at constant speed.

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**The vertical component of a projectile is same as for an object in free fall.**

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**The vertical motion of a horizontally fired projectile is the same as free fall.**

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For a projectile fired at an angle, the projectile will be 5t2 below where it would be if there was no gravity.

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