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The AP Annual National Conference Las Vegas, Nevada July 19, 2013 Ahmed Salama PANTHER Academy, Paterson P.C.C.C. College [email protected] 1 AMTNJ Annual Conference October 24-25, 2014 Use trigonometric functions and calculus skills to determine the range and final velocity of a projectile based in knowing its initial velocity and maximum height. This activity provides a good example of connecting physics and mathematics. 3 Mathematically, how the projectile works !! 4 Rocket 2: 60 Rocket 1: Range VS Theoretical Range at 40˚ 60 40 40 30 Range Distance 50 Distance 50 Range VS Theoretical Range at 40˚ 30 Range Theoretical Range Theoretical Range 20 20 10 10 0 0 40 60 80 Pressure 100 40 60 80 Pressure Compare rocket 1 to rocket 2 in range and theoretical range at 400 angle Rocket 1= .073 kg and Rocket 2 = .086 kg 100 Rocket 2: Rocket 1: Range VS Theoretical Range at 45˚ 60 Range VS Theoretical Range at 45˚ 60 50 50 40 30 Range Theoretical Range Distance Distance 40 30 Theoretical Range 20 20 10 10 0 0 40 60 80 Pressure 100 Range 40 60 80 Pressure Compare rocket 1 to rocket 2 in range and theoretical range at 450 angle Rocket 1= .073 kg and Rocket 2 = .086 kg 100 Rocket 1: Rocket 2: Range VS Theoretical Range at 50˚ Range VS Theoretical Range at 50˚ 60 45 40 50 35 40 25 Range 20 Theoretical Range 15 10 Distance Distance 30 Range 30 Theoretical Range 20 10 5 0 0 40 60 80 Pressure 100 40 60 80 Pressure Compare rocket 1 to rocket 2 in range and theoretical range at 500 angle Rocket 1= .073 kg and Rocket 2 = .086 kg 100 0 sin 0 0 − R Students Real Experiment Example 1: An air compressor system projectile is fired at 45 0 with initial velocity 14.16 m/sec. it took 3.4 seconds to reach the ground. Investigate the maximum range in theory and compare it to the reality. 2 x v 0 cos x 14 . 16 cos 45 0 t , y v 0 sin t 1 gt 2 2 1 0 2 ( 3 . 4 ), y 14 . 16 (sin 45 )(1 . 7 ) ( 9 . 8 )(1 . 7 ) 2 x =14.16(0.7)(3.4) , X= 34.04 m ( the range ) Y= 14.16(0.7)(1.7)- 0.5(9.8)(1.72) y= 2.86 m the maximum height In the case of y, we substitute for1 one half of the time period to find the maximum height. 2 t 9 Students measured the range and it was 16.10 meters which is different totally than their computation 34.04 meter. Students were looking for explaining to the resulting error which was -17.94 They came up with Air resistant as a reason for having different reading between their math work and the experiment reading. They set up this equation for horizontal velocities that oppose wind speed. v x v 0 cos v R Such as vR is the wind velocity By taking the derivative of both sides they will get the range formula x dv dt v 0 (cos ) t v R t Students have researched about the wind speed and direction at Paterson, NJ See this link : http://www.friendlyforecast.com/usa/ archive/archive.php?region=NJ&id=204 500&?-Forecast-Paterson-NewJersey&date=20130524000000&sort=h our The speed wind readings were various among 8.1 mph and 6.9mph and 4.6 m/h between time 12PM to 4:00 PM . The students estimated the average which is 7 m/h or 3.1 m/s NW We were launching projectile towards North so students decided to multiply speed of Air by Cos (450) The resulting range is X= (14.16 (cos 45o)(3.4))- ((5 cos 450)(3.4)) = 22.02 m it is closer to the reality but with error of -5.92 so our result is still not accurate Why ? Physics turn is ended , Calculus starts to tell us the fact of having errors in our work. Now , we need to search about the affect of drag coefficient and Air Resistant. Assume that our projectile is launched at t =0 making an angle ө to the horizontal. In addition to the force of gravity, the projectile is subject to an air resistance, We get some idea of how air resistance and drag force modify projectile trajectories. The equation of projectile`s motion is F = mg – cv Such as c is constant of drag and resistant force, it depends on the size of projectile and its mass. a is gravity acceleration 9.8 m/s , v is the velocity , m is the rocket mass. ma = mg- cv , then m dv dt mg cv Using differential equation to separate parts: dv x g dt vx vt and vy g 1 dt v t dv y Students have learned that terminal velocity vt vt 2w c d pA Is affected by the chemistry of the atmosphere and the gravitational constant of the earth or any planet . It is velocity that rocket can not accelerate after it. Terminal velocity - the constant maximum velocity reached by a body falling through the atmosphere under the attraction of gravity v x ln v x0 g t vt And take e for both sides gt v x v 0 cos e x v 0 v t cos g vt 1 e Another integration to both sides gt / v t Stop for a moment and go back to physics Calculate the terminal velocity W is weigh , cd is drag coefficient , P is pressure A is cross sectional area of projectile vt 2w c d pA vt m 2 mg c d pA Is mass of our rocket 7.3gram = .0073 Kg P is the density of air (p = 1.29 kg/m^3) A is cross sectional area = =.01272 , Where r= .5 inch = .0127 meter Cd is drag coefficient depends on the mass Let is calculate our projectile `s drag coefficient. Net force = Drag – Weight = 0 F 1 2 c d pAv 2 mg 0 Solve it for cd is from 4 and 1.13 it is calculated by Mr. Oettinger science teacher Substitute for all values vt 2 mg c d pA 2 . 0073 9 . 8 vt 1 . 13 1 . 29 . 0127 Vt = 13.9 m/s2 2 Back again to maximum range : x v 0 v t cos g 1 e gt / v t Substitute for all values x (14 . 16 )(13 . 92 ) cos 45 9 .8 0 1 e ( 9 . 8 )( 3 . 4 ) / 13 . 92 X= 12.92 meter which is matched to the actual experiment reading with error of 3 m Go to web-site projectile motion animation: http://phet.colorado.edu/sims/ projectile-motion/projectilemotion_en.html A baseball is hit from 2.8 feet above the ground with an initial velocity of 160 ft/sec o at an angle of 25 from the horizontal. A gust of wind adds a component of -9 ft/sec in the horizontal direction to the initial velocity. 25 The Parametric Equations become: 1 X= ( V0cos Ө-VR)t , y = y0+(v0sinӨ)t- gt 2 2 Given: v = 160ft/s , y = 2.8 ft, g = 32 ft/s 0 1 2 0 g 2 2 16 ft / s 2 25 ِِِWe could find the time by using the formula: t 0 v 0 sin t g 160 sin 25 0 2 . 1 sec 32 The time to reach the ground is 2t = 4.2 sec 26 The Range: X= (160cos(25) -9)4.2 = 285.6 ft The Maximum Height (Altitude): y= 2.8+(160sin(25 )(2.1))-(0.5(32)(2.1) 2 ) y= 74.2 ft 27 These equations can be modeled graphically by using Ti- 84 to model the path of the ball: First set your calculator on Parametric Version and Degree System. 28 Type your parametric equations and adjust your window: 29 Graph the equations: 30 Press trace to find out the maximum height Is 74.2 ft will be at the time 2.1 second. 31 And the Range is 584.8 ft at time 4.3 seconds or 4.2 seconds. 32 Example 3 Ball tossed with 10 m/s vertical velocity from window 20 m above ground. Determine: • velocity and elevation above ground at time t, • highest elevation reached by ball and corresponding time, and • time when ball will hit the ground and corresponding velocity. 33 SOLUTION: • Integrate twice to find v(t) and y(t). dv a 9 . 81 m s dt v t t dv 9 . 81 dt v0 2 v t v 0 9 . 81 t 0 m v t 10 9 . 81 t 2 s s m dy v 10 9 . 81 t dt y t t dy 10 9 . 81 t dt y0 y t y 0 10 t 1 9 . 81 t 2 0 m 2 m y t 20 m 10 t 4 . 905 t 2 s s 34 2 • Solve for t at which velocity equals zero and evaluate corresponding altitude. m v t 10 9 . 81 t 0 t 1 . 019 s 2 s s m m m 2 y t 20 m 10 t 4 . 905 t 2 s s m m 2 y 20 m 10 1 . 019 s 4 . 905 2 1 . 019 s s s y 25 . 1 m 35 • Solve for t at which altitude equals zero and evaluate corresponding velocity. m 2 m y t 20 m 10 t 4 . 905 t 0 2 t 1 . 243 s meaningles s s t 3 . 28 s s m v t 10 9 . 81 t 2 s s m m v 3 . 28 s 10 9 . 81 3 . 28 s 2 s s m v 22 . 2 m s 36 • You wish to fire an anti-aircraft shell to intercept an enemy plane flying towards you at 600 mph and an altitude of 42,000ft. If the plane is initially 20 miles away, and your artillery piece has a muzzle velocity of 2000 ft/sec, how long do you have to adjust the piece to an angle of 600 and fire? 37 38 y a vo 60 x 39 y0 =0 V0y , = vo sin 60 = ay = - 32 ft/s2 . (.866)(2000) = 1800 ft/sec. The general equations of motion for constant acceleration in 2-dimensions are: x (t ) 1 y (t ) 1 a t 2 x a 2 v v x y y t 2 2 v 0x v 0 t x 0 yt y 0 (t ) a t v 0 x x (t ) a y t v0 y 40 We insert the known values for acceleration & initial conditions and obtain the specific equations for the2 shell: x(t) = 1000 t ; y(t) = - (1/2)(32)t + 1800 t vy(t) =- 32 t + 1800 . We can now answer any question regarding the motion of the shell. One question that might occur to us is: Can we hit a plane at 42,000 ft.? Let's determine the maximum height to which the shell will rise. 41 vy(t') = 0 = -32 t' + 1800 , t' = 56.3 sec. 2 Then: ymax = y(t') = -(1/2)(32)(56.3) + 1800 (56.3) = 50,625 ft. We can reach the elevation of the plane. The next question is: When will the shell be at the plane's elevation (42,000 ft)? 2 t 2 y(t') = 42,000 = - 16 t2 + 1800 t we could solve it for y by using graphing calculator. 42 43 44 By using the graphing calculator we have solved the quadratic equation: 42,000 = - 16 t2 + 1800 t . The graphing process yields the results: t1 = 33.03 sec; t2 = 79.47 sec. 45 At the first time the shell will be a distance x (t'=33.03) = 33,030 ft (6.256 mi) from where it was fired. At the second time x (t'= 79.47) = 79,470 ft (15.05 mi). But the plane traveling at 880 ft/sec will cover a distance of 13.245 mi in 79.47 seconds. Hence, we can't hit the plane with the shell on the way down. 46 We illustrate in the figure the various distances involved. xh = 6.256 mi is position where we expect to hit the plane; xo = 20 mi is initial position; xf = ?? is position of plane when we fire the shell. Since the plane will travel a distance of (880)(33.03)/(5280) = 5.505 mi while the shell is in the air, then xf = 6.256 + 5.505 = 11.76 mi away. 47 y vo xh xf x o 48 Now the question is, "How long do we wait till the plane reaches xf? Taking the difference (20 11.76) and dividing by the planes speed, gives us 49.44 seconds. There is one other 'minor' question we should ask ourselves. Where will the plane drop its bomb??? If it will release its bomb at a distance greater than the 11.76 miles then we better get the heck out of here!!! A falling bomb will have equations of motion given by: (using CS with y=0 the ground). 49 xB(t) = 880 t yB(t) = -(1/2)(32) t2 +42,000 . Hence, the time to 'fall" is: yB(t') = 0 = -16t2 + 42,000 t' = 51.23 sec. In this time, the bomb travels a horizontal distance of (880)(51.23)/5280 = 8.54 mi. Thus the plane plans to release its bomb when it is 8.54 miles from us. But we know the plane will never reach this point. It will be shot down 11.76 miles away! 50 Particle and projectile in Calculus test: Try : hand out sheet: Summary: We have presented the following: 1) Real Class Experiment to study the affect of air resistance and drag force on the accuracy of projectile. 2) The perfect elevation angle of launching rocket with and without air resistance. 3) Use differential equation to compute terminal velocity , drag coefficient and terminal velocity. 4) Baseball motion and time to reach the ground under the condition of baseball playground. (sport application). 5) Ball tossed with vertical velocity from tower window. 6) Firing an anti-aircraft shell to intercept an enemy plane flying towards you(Military application). 7) AP- Calculus sample problems. Citation: http://www.friendlyforecast.com/usa/archive/archive.php?region=NJ&id=204500&?Forecast-Paterson-New-Jersey&date=20130524000000&sort=hour http://www.grc.nasa.gov/WWW/k-12/rocket/termvr.html http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node29.html http://demonstrations.wolfram.com/ProjectileWithAirDrag/ http://www.apogeerockets.com/downloads/Newsletter303_Large.pdf http://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.html http://www.rocketmime.com/rockets/rckt_eqn.html Books resources: AP Calculus ninth edition for: Ron Larson, The Pennsylvania State University. Bruce H. Edwards, University of Florida The physics Book for Clifford A. Pickovee Administration and support Mr. Joseph Oettinger , science review Ms. Florita Cotto, Vice principal Sharo Hilda , Principal Maria Santa, Interim Superintendent Jacqueline Jones, Chief Staff Supervision Janice Basilicato, Interim Director of Math Department. Barbara Rugys, Math supervisor k-12 Dr David Rust, Math Chairperson