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Two Dimensional Motion Two components: Horizontal (x-axis) & Vertical (y-axis)
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Ex:1) projectile fired from a gun 2) car turning around a corner 3) child swinging on a playground
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Independence of Motions Vector Components taken perpendicular to each other are independent. ***horizontal & vertical motions are independent***
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If you fired a gun horizontally and dropped a similar bullet from the same height, at the same time as the other bullet left the barrel, which bullet would hit the ground first?
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Ex- a bullet fired from a gun falls at the same rate & hits the ground at the same time as a similar bullet simply dropped from the muzzle
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Projectiles Objects that are thrown or launched into the air and are subject to gravity.
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Trajectory A curved path, called a parabola, that a projectile follows.
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Air resistance slows down projectile, as it collides with air molecules. Thereby diminishing the trajectory.
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We are going to neglect the effects of air resistance
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Projectiles fired horizontally.
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Ex: Let’s analyze a bullet fired from a gun from a height of 80 meters with an initial velocity of 200 m/s.
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Horizontal: 1 st the gun exerts a force on the bullet, giving it motion - let v i,x = 200 m/s
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Are there any horizontal forces to change the bullets speed once it is shot? NO!
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remains constant
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Vertical: = 0 m/s bullet fired horizontally What vertical forces act on the bullet once it is shot? Gravity = - 10 m/s 2
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What happens to the bullet’s velocity as it falls? Increases negatively with time due to gravity
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Recall frame of reference: To locate bullet at any time, t, after it is shot…
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To figure out the displacement Horizontal ( x): there is no acceleration once the bullet leaves the barrel.
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Once the projectile is launched, it no longer undergoes acceleration. Therefore, the horizontal velocity (v x ) remains constant.
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v x = v i,x = constant
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Vertical ( y): when the bullet leaves the barrel there is no vertical velocity initially, but gravity takes right a hold of it.
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acceleration is gravity
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v f,y = g t v f,y 2 = 2 g y
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time x = v xi t y = ½ g t 2 y = height+ y (s) (m) (m) (m) 0 1 2 3 4 0 200 400 600 800 0 - 5 - 20 - 45 - 80 80 75 60 35 0
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Projectiles Fired Horizontally
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Projectiles fired horizontally at different speeds
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Ex : Joe kicks a 0.75 kg ball off a 50 m cliff. If he gives the ball only an initial horizontal speed of 15 m/s. A) How long does it take to hit the ground? B) how far does it travel in the horizontal direction?
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G: m =0.75 kg, y= -50 m v i,x =15 m/s, g= -10 m/s 2 U: t = ?, x =?
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G: m =0.75 kg,v i,x =15m/s, g= - 10 m/s 2, y= - 50 m U: t = ?, x =? E: A)
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S:
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E: B) x = v i,x t S: x = (15 m/s) x (3.16 s) S: x = 47.4 m
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Ex 2: A golf ball is hit off a cliff at a speed of 80 m/s horizontally and travels 471 m. A) Calculate the height of the cliff.
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G: g = -10 m/s 2, x = 471 m, v i,x = 80 m/s U: y = ?
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G: g = - 10 m/s 2, x = 471 m, v i,x =80 m/s U: y=? E:
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G: g = -10 m/s 2, x = 471 m v i,x = 80 m/s U: y = ? E: Need to find t
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t = x / v i,x t = 471 m / 80 m/s t = 5.89 sec
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S: y = ½ (-10)(5.89) 2 S: y = -173.5 m
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B) How fast will the ball hit the ground? Since the ball is moving in both the x and y directions we need to combine both velocities.
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Find v f,y v f,y = g t v f,y =(-10 m/s 2 )(5.89 s) v f,y = - 58.9 m/s
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E:
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S:
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E: S:
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Projectile Fired at an Angle When an object is thrown up or at an angle, the initial y- component velocity decreases due to gravity.
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At its peak the ‘y’ velocity is zero, acceleration is still –10 m/s 2. The object begins to fall due to gravity, increasing its speed, but NEGATIVE.
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When it reaches the point where it was launched from, its velocity is the same as the initial y component, but negative.
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Ex 3: Consider a cannon firing at a 30 o angle with an initial speed of 39.2 m/s.
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Remember to consider independent components separately!
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Horizontal Formulas :
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From vectors, how did we find the x component of vector at an angle?
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Horizontal Formulas :
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Remember horizontal motion is constant.
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Recall, no forces change horizontal motion …therefore
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Vertical:
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Formulas : From vectors, how do we find the y component of a vector at an angle?
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Vertical: Formulas :
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Vertical: Formulas :
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v f,y = v i (sin )+g t v f,y 2 = v i 2 (sin ) 2 + 2g y
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The cannon ball decelerates to a maximum height where and then accelerates to the ground.
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Horizontal Distance Traveled by a Projectile Range- horizontal distance traveled by a projectile
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-Maximum range achieved with a firing angle of 45 degrees.
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Actual Motion of a Projectile Simplifying assumptions: 1)Air resistance can be neglected 2)Earth can be considered flat for small ranges
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3) For short ranges, gravity acts in the same direction constantly 4) Projectiles remain in the same vertical plane throughout flight
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Effect of these assumptions: projectile turns downward from it’s ideal path Thus, it has a shorter range
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Monkey and the Hunter
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http://www.phys.unsw.edu.au/~jw/d emo/projectiles.html http://www.phy.ntnu.edu.tw/ntnujava/index. php?topic=144.0http://www.phy.ntnu.edu.tw/ntnujava/index. php?topic=144.0
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http://www.phys.unsw.edu.au/~jw/d emo/projectiles.html http://www.phy.ntnu.edu.tw/ntnujava/index. php?topic=144.0http://www.phy.ntnu.edu.tw/ntnujava/index. php?topic=144.0
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When a projectile lands or is caught at the same height as it was launched: 1.The vertical displacement is zero. y= 0 m 2.The final ‘y’ velocity is the negative of the initial ‘y’ velocity. v f,y = -v i,y
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If you are given the initial velocity @ an angle always find the x and y components of the velocity.
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Ex 4: A baseball is hit 45 m/s at 45 o. A) How long is the ball in the air? B) How far does the ball travel horizontally?
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G: v i = 45 m/s, =45 o g = -10 m/s 2 U: t = ?, x =? Find the ‘x’&‘y’ velocity components first.
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Since the angle is 45 o both x and y components are the same. (forms an isosceles triangle)
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E: v i,x = v i cos S: v i,x = 45cos(45) S: v i,x = 31.8 m/s v i,x = v i,y = 31.8 m/s
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E:
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Remember the final ‘y’ velocity is the negative of the initial ‘y’ velocity.
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S:
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S: t = 6.36 sec
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E:
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E: S:
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E: S:
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EX 5: In a scene from an action movie a stunt man must leap from one building to another that is 4 m apart. After a running start the actor leaps from the roof at an angle of 15 o with a speed of 5 m/s. Will he make the it to the other roof, which is 2.5 m shorter than the one he jumped from?
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G: y = - 2.5 m, g = -10m/s 2, v i = 5 m/s, = 15 o U:
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass Increase Velocity Increase Angle Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increase Angle Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increase Angle Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increase Angle Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increase Angle Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increases Increase Angle Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increases Increase Angle Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increases Increase Angle Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Increases Note 4 Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Increases Note 4 Stays the Same Add Air Resistance
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Increases Note 4 Stays the Same Add Air Resistance Decreases
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Increases Note 4 Stays the Same Add Air Resistance Decreases
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Increases Note 4 Stays the Same Add Air Resistance Decreases
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When you : Range ( x) Height ( y) Time ( t) Final Velocity Increase Mass No Change Increase Velocity Increases Note 1 Increase Angle Note 2 Increases Note 3 Increases Note 4 Stays the Same Add Air Resistance Decreases
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Note 1 As initial velocity changes, the final velocity follows. (It’s the same)
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Note 2 Increasing the launch angle up to 45 degrees increases the range.
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After 45 degrees, the range decreases. 45 degrees gives you the maximum range.
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Complimentary angles (i.e. 15 & 75) give you the same range.
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90 degrees gives you the maximum height. Note 3
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Note 4 90 degrees gives you the maximum ‘hang’ time.
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