Presentation on theme: "Do now A B + = ? The wrong diagrams Draw the right diagram for A + B."— Presentation transcript:
1 do nowAB+= ?The wrong diagramsDraw the right diagram for A + B
2 3.3 projectile motionObjectives 1. Recognize examples of projectile motion. 2. Describe the path of a projectile as a parabola. 3. Resolve vectors into their components and apply the kinematic equations to solve problems involving projectile motion.
3 questionThe long jumper builds up speed in the x-direction and jumps, so there is also a component of speed in the y direction. Does the angle of take-off matter to the jumper?
4 A parabola is the set of all points in the plane equidistant from a given line (the conic section directrix) and a given point not on the line (the focus).y ~ x2
5 What is a projectile?An object that is launched into the air with some INITIAL VELOCITYCan be launched at ANY ANGLEIn FREEFALL after launch (no outside forces except force of gravity)The path of the projectile is a PARABOLA
11 Horizontal Projectiles Over the EdgeHorizontal Projectiles
12 Projectiles Launched Horizontally Vertical motion:Free fall free rest:vy = 0ay = m/s2Horizontal motion:vx = vix = constantax = 0Thus, the projectile travels with a constant horizontal velocity and a downward vertical acceleration.
13 Velocity of Horizontal Projectiles Horizontal motion is constant: velocity is constant.Vertical: same as drop the ball from rest: velocity is increasing by 9.81 m/s every second
14 Projectile launched horizontally A projectile is any object upon which the only force is gravityProjectiles travel with a parabolic trajectory due to the influence of gravity,In horizontal direction, the projectile has no acceleration, its velocity is constant: vx = viIn vertical direction, the projectile has acceleration: a = m/s/s. Its initial vertical velocity is zero and it changes by m/s each second. Same as an object falling from rest.The horizontal motion of a projectile is independent of its vertical motion
15 ExampleAn object was projected horizontally from a tall cliff. The diagram represents the path of the object, neglecting friction.Comparing the following at point A & B:AccelerationHorizontal velocityVertical velocity
16 As the red ball rolls off the edge, a green ball is dropped from rest from the sameheight at the same timeWhich one will hit the ground first?They will hitat the SAMETIME!!!
17 The same time?!? How?!? The green ball falls from rest and has no initialvelocity IN EITHERDIRECTION!viy and vix = 0The red ball has an initialHORIZONTAL velocity (vix)But does not have any initialVERTICAL velocity (viy = 0)vix
18 We can find an object’s displacement in EITHER DIMENSION using TIME One Dimension at a TimeBoth balls begin with no VERTICAL VELOCITYBoth fall the same DISTANCE VERTICALLYFind time of flight by solving in the appropriate dimensionWe can find an object’s displacement inEITHER DIMENSION using TIME
19 Example #1A bullet is fired horizontally from a gun that is 1.7 meters above the ground with a velocity of 55 meters per second.At the same time that the bullet is fired, the shooter drops an identical bullet from the same height.Which bullet hits the ground first?Both hit the ground at the same time
20 Equation for horizontal projectile θ = 0Verticalay = m/s2Horizontalax = 0d = ½ (vi + vf)tvf = vi + atd = vit + ½at2vf2 = vi2 + 2adviy = visinθ = 0y = ½ (viy + vfy)tvfy = viy + ayty = viyt + ½ayt2vfy2 = viy2 + 2ayyvix = vicosθ = vix = vix∙tx and y has the same t
21 Need time from vertical Example #2An airplane making a supply drop to troops behind enemy lines is flying with a speed of 300 meters per second at an altitude of 300 meters.How far from the drop zone should the aircraft drop the supplies?Need time from verticaldy = viyt + ½ ayt2300 m = 0 + ½ (-9.81 m/s2)t2t = 7.82 sUse time in horizontaldx = vixt + ½ axt2dx = (300 m/s)(7.82 s) + 0dx = 2346 m
22 Need time from vertical Example #3A stuntman jumps off the edge of a 45 meter tall building to an air mattress that has been placed on the street below at 15 meters from the edge of the building.What minimum initial velocity does he need in order to make it onto the air mattress?Need time from verticaldy = viyt + ½ ayt245 m = 0 + ½ (-9.81 m/s2)t2t = 3.03 sUse time to find vv = d / tv = 15 m / 3.03 sv = 4.95 m /s
23 Need time from vertical Example #4Example #4A CSI detective investigating an accident scene finds a car that has flown off the edge of a cliff. The car is 79 meters from the edge of the 25 meter high cliff.What was the car’s initial horizontal velocity as it went off the edge?Need time from verticaldy = viyt + ½ ayt225 m = 0 + ½ (-9.81 m/s2)t2t = 2.26 sUse time in horizontaldx = vixt + ½ axt279 m = vix (2.26 s) + 0vix = m/s
24 Example #5The path of a stunt car driven horizontally off a cliff is represented in the diagram below. After leaving the cliff, the car falls freely to point A in 0.50 second and to point B in 1.00 second.Determine the magnitude of the horizontal component of the velocity of the car at point B. [Neglect friction.]Determine the magnitude of the vertical velocity of the car at point A.Calculate the magnitude of the vertical displacement, dy, of the car from point A to point B. [Neglect friction.]
25 Class work Page 101 - Sample problem 3D Page 102 – practice 3D Answers:0.66 m/s4.9 m/s7.6 m/s5.6 m
30 Initial velocity 12 m/s 30° Pythagorean Theorem vi2 = vix2 + viy2 What is the vertical part ofthe soccer ball’s initial velocity?What is the horizontal part ofthe soccer ball’s initial velocity?viy = vi sin θvix = vi cos θ12 m/s6 m/s30°10.4 m/s
31 What do we know or assume about the vertical part of a projectile problem? Initial vertical velocity = vi sin θAcceleration = m/s2Vertical speed will be 0 at the maximum heightTime to top = HALF total time in the airFind time to top using final velocity equationVertical Distance – Max heightUse time to top and solve vertical distance equation
32 What do we know or assume about the horizontal part of a projectile problem? Initial vertical velocity = vi cos θAcceleration = 0 (if we assume no air resistance)Horizontal Distance – RangeUse total time and solve horizontal distance equation
33 The symmetrical nature of a ground launched projectile What is the acceleration at the top of the path?What is the vertical velocity at the top of the path?
34 Projectile launched at an angle A projectile is any object upon which the only force is gravityProjectiles travel with a parabolic trajectory due to the influence of gravity,In horizontal direction, the projectile has no acceleration, its velocity is constant: vx = vicosθIn vertical direction, the projectile has acceleration: a = m/s/s. Its velocity of a projectile changes by m/s each second. Same as a free falling object.The horizontal motion of a projectile is independent of its vertical motion
35 A projectile is fired with initial horizontal velocity at 10 A projectile is fired with initial horizontal velocity at m/s, and vertical velocity at +20. m/s. Determine the horizontal and vertical velocity at 1 – 5 seconds after the projectile is fired. Use g = 10 m/s/s.
36 TimeHorizontalVelocityVerticalAccelartion0 s1 s2 s3 s4 s5 s
38 Initial Velocity Components Since velocity is a vector quantity, vector resolution is used to determine the components of velocity.vi2 = vix2 + viy2SOH CAH TOAsinθ = viy / viviy = visinθcosθ = vix / vivix = vicosθviviyθvixSpecial case: horizontally launched projectile:θ = 0o: viy = visinθ = 0; vix = vicosθ = vi
39 Examples Determine the horizontal and vertical components A water balloon is launched with a speed of 40 m/s at an angle of 60 degrees to the horizontal.A motorcycle stunt person traveling 70 mi/hr jumps off a ramp parallel to the horizontal.A springboard diver jumps with a velocity of 10 m/s at an angle of 80 degrees to the horizontal.
40 exampleA machine fired several projectiles at the same angle, θ, above the horizontal. Each projectile was fired with a different initial velocity, vi. The graph below represents the relationship between the magnitude of the initial vertical velocity, viy, and the magnitude of the corresponding initial velocity, vi, of these projectiles. Calculate the magnitude of the initial horizontal velocity of the projectile, vix, when the magnitude of its initial velocity, vi, was 40. meters per second.PropertiesReasonAnswerHint
41 The point of resolving an initial velocity vector into its two components is to use the values of these two components to analyze a projectile's motion and determine such parameters asthe horizontal displacement,the vertical displacement,the final vertical velocity,the time to reach the peak of the trajectory,the time to fall to the ground, etc.
42 ExampleA football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football.Horizontal ComponentVertical Componentax = 0 m/s/svix = 25 m/s•cos45o= 17.7 m/sx = ?ay = m/s/sviy=25 m/s•sin45o= 17.7 m/svfy = m/symax = ?ttotal = ?
43 Solve for t – use vertical information: vfy = viy + aytt = 3.61 sSolve for x – use horizontal information:x = vix•t + ½ •ax•t2x = 63.8 mSolve for peak height – use vertical information:vfy2 = viy2 + 2ayypeakAt the top, vfy = 0ypeak = 15.9 m
44 exampleA long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.Horizontal ComponentVertical Componentax = 0 m/s/svix = vi•cosθvix = 12 m/s•cos28ovix = 10.6 m/sttotal = 2tupx = ?ay = -9.8 m/s/sviy = vi•sinθviy = 12 m/s•sin28oviy = 5.6 m/svfy = -5.6 m/sy = ?tup = ?t = 1.1 sx = 12.2 mypeak = 1.6 m
45 ExampleA cannon elevated at an angle of 35° to the horizontal fires a cannonball, which travels the path shown in the diagram. [Neglect air resistance and assume the ball lands at the same height above the ground from which it was launched.] If the ball lands 7.0 × 102 meters from the cannon 7.0 seconds after it was fired,what is the horizontal component of its initial velocity?what is the vertical component of its initial velocity?
46 An object is thrown horizontally off a cliff with an initial velocity of 5.0 meters per second. The object strikes the ground 3.0 seconds later. What is the vertical speed of the object as it reaches the ground? [Neglect friction.]
47 In the diagram, a 10.-kilogram sphere, A, is projected horizontally with a velocity of 30. meters per second due east from a height of 20. meters above level ground. At the same instant, a 20.-kilogram sphere, B , is projected horizontally with a velocity of 10. meters per second due west from a height of 80. meters above level ground. [Neglect air friction.] Initially, the spheres are separated by a horizontal distance of 100. meters. What is the horizontal separation of the spheres at the end of 1.5 seconds?
48 Total flight time, range, max height Time to go upt = visinθ / gAs θ increases, flight time increase Max time: θ = 90oRangeRange = vi2sin2θ /gProjectile has maximum range when θ = 45oMax heighthmax = (visinθ)2/2gAs θ increases, flight height increase Max height: θ = 90o
49 Class work Page 103 – sample problem 3E Page 104 – practice 3E #1-5 Answers:Yes70.3 m20. s; 4.8 m6.2 m/s17.7 m/s; 6.60 m