Projectile Motion notes.notebook February 25, 2014 Projectile Motion thephysicsclassroom.com Mar 72:18 PM Projectile Motion Now we have 2 dimensions and we break the initial velocity into its 2 components, horizontal and vertical the ball is kicked Initial Velocity as 2 dimensions/components Mar 72:50 PM 1 Projectile Motion notes.notebook February 25, 2014 IMPORTANT ; g only acts vertically, so the vertical velocity will change horizontal velocity remains constant. Mar 73:04 PM if the initial velocity is 20 m/s at an angle of 15 degrees what are the horizontal and vertical components? horizontal componenet 20 cos 15 = 19 m/s vertical component 20 sin 15 = 5.2 m/s Mar 72:50 PM 2 Projectile Motion notes.notebook February 25, 2014 Once we have the components: horizontal (no acceleration) use use the 4 equations for UARM vertical (a = -9.8 m/s2) The only variable they share is time Once you have the components DONT USE THE INITIAL VELOCITY PUT A LINE THROUGH IT AND ONLY USE THE COMPONENTS VX FOR HORIZONTAL VY FOR VERTICAL Mar 72:57 PM Let's start with an example of a ball that is launched with a velocity of 1.2 m/s at an angle of 400 How high will it go? How far will it go? Height is vertical and length is horizontal Divide the paper in half: find the components only share time vertical a = -9.8 m/s2 use 4 equations UARM horizontal no accelertion v = d/t Mar 72:59 PM 3 Projectile Motion notes.notebook February 25, 2014 Feb 2511:36 AM Horizontal Launch (the angle is 00) That means, for the y component vi = 0 m/s and all the velocity is horizontal at t = 0s. A marble rolls off the table with a horizontal velocity of 1.2m/s It lands in a cup 0.51m from the table's edge. How high is the table? Horizontal vi = 1.2 m/s vf = 1.2 m/s di = 0m df = 0.51m knowing the velocity is constant I know that v = d/t so I can find t t = d/v = 0.51m/ (1.2m/s) = 0.43s Now I have time Vertical vi = 0 m/s g = -9.8 m/s2 di = x df = 0m t = 0.43s What forumal? df = di + vit + 1/2at2 0m = x + 0*t -4.9t2 x = 4.9*(.43)2 x = 0.91m The table is 0.91 m high Mar 72:59 PM 4 Projectile Motion notes.notebook February 25, 2014 Helpful Hints: It is common to put the origin of the cartesian plane at ground level where x = 0 is at the launch site Projectile objects leaving the ground will reach the peak height at half the total time If taking off from a different level (height), the time it takes to return to it's original height is twice the time it takes to reach the peak height Mar 73:19 PM Emmanuel Zacchini, the famous human cannonball, was fired out of a cannon with a speed of 24.0 m/s at an angle of 40.0 degrees to the horizontal. If he landed in a net 56.6 m away at the same height from which he was fired, how long was Zacchini in the air? Find the horizontal speed, and given the distance find t. Vh = 24.0 m/s cos 40 = 18.4 m/s t = d/v (velocity is constant) = 56.6m/ 18.4 m/s =3.08s Mar 72:20 PM 5 Projectile Motion notes.notebook February 25, 2014 Evel Knievel, successfully jumped 69.5 m over a Grand Canyon gorge. Assuming that he started and landed at the same level and was airborne for 3.66s, what height from his starting point did this daredevil achieve? Knowing the max height was at the halfway point, we know that t = 1.83s g= -9.8m/s2 What equation df = di + vit + 1/2at2 If we say vi = 0 at the halfway point then di= x and df = 0 Then 0 = x + 1/2(-9.8m/s2)(1.83s)2 x = 16.4m so the motorcycle reached a height of 16.4m Mar 73:28 PM Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly for lunch. If the lily pads are spaced 2.4 m apart, and Ferdinand jumps with a speed of 5.0m/s, taking 0.60 s to go from lily pad to lily pad, at what angle must Ferdinand make each of his jumps? df= 2.4 di = 0m vi = 5.0m/s t = 0.60s horizontal v =d/t 2.4/0.69 = 4 m/s The horizontal component of vi = vicosθ = vx 5.0 m/s * cos θ = 4 m/s cos θ = 4/5 θ = cos-1(4/5) = 37 degrees to horizontal Mar 73:36 PM 6 Projectile Motion notes.notebook February 25, 2014 We often find the time first and then plug the time in to find other variables. Let's look at the examples in the textbook p 252-253 A soccer player kicks the ball with a velocity of 20.0 m/s at a 25.0 degree angle above the horizontal How long does it take for the ball to reach its maximum height? Mar 73:43 PM A soccer player kicks the ball with a velocity of 20.0 m/s at a 25.0 degree angle above the horizontal How long does it take for the ball to reach its maximum height? vix = 20.0 cos 25.0 v is constant viy = 20.0 sin 25.0 v2 = 0m/x What formula? #1 Vf = Vi + at 0m/s = 8.45m/s - 9.8t -8.45/-9.8 = t t = 0.862s Mar 73:43 PM 7 Projectile Motion notes.notebook February 25, 2014 What is the balls flight time? We know it should be 2 * 0.86 seconds or 1.72 seconds We can also use the formula yf = yi + vit +1/2at2 Yf will be 0 again (it is back on the ground) 0 = 0 + 8.45m/s t - 4.9 m/s2t2 0 = t(8.45 -4.9t) t = 0 (when the ball was kicked 9.45/4.9 = t t= 1.72s Mar 73:51 PM What distance does the ball travel horizontally? We know the time is 1.72s The horizontal velocity is 20.0m/s cos 25.0 = 18.1m/s (and its constant so vi and vf = 18.1m/s) v = d/t and d = vt so d = 18.1m/s*1/72s = 31.2m Mar 73:56 PM 8 Projectile Motion notes.notebook February 25, 2014 Example B: Here is the question: Try it and check your answer in the text book. From a roof 50.0m high, a ball is thrown with a velocity of 5.00m/s at 25.0o above the horizontal. List all your knowns for vertical and horizontal: a) how long does it take for the ball to reach the ground? ( hint: you will need to use the quadratic equation: use the positive value of t) b) what is the magnitude of the ball's velocity at the moment it touches the ground? What is the direction of the velocity vector at this moment c) what is the maximum height reached by the ball? Mar 75:45 PM How do we get the resultand vector and the angle? θ 8.45m/s 18.1m/s Pythagorean theorem: find the hypotenuse = 20.0m/s tan θ = -8.45/181 θ = -250 3600 - 250 = 3350 from the horizontal. Mar 75:34 PM 9 Projectile Motion notes.notebook February 25, 2014 Problems: section 11.2, 251 all Next time; design your own lab...... What factors influence projectile motion? Mar 75:51 PM d) what is the value of the ball's velocity at the moment it touches the ground? give the magnitude and direction.... (yes it is a vector with an x component and a y component) we know vxf = 18.1 m/s vyf = ? we have a and t and vi so we can use: vf = vi + at = 8.45m/s + (-9.8m/s2)(1.72s) = -8.45m/s (directed down) 8.45m/s 18.1m/s How do we get the resultand vector and the angle? Mar 75:34 PM 10 Projectile Motion notes.notebook February 25, 2014 Mar 73:47 PM 11
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