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PHYSICS: Vectors and Projectile Motion Day 1 Today’s Goals Students will: 1. Be able to describe the difference between a vector and a scalar. 2. Be able to draw and add vector’s graphically and use Pythagorean theorem when applicable. A couple of quick definitions! This should be review, but just in case….. • Scalars • Vectors Remember: Force, Velocity, Displacement, Acceleration are ALL VECTORS!!! Representing Vectors • Because of direction, vectors do not add the same way scalars do. 2 + 2 does not always equal 4!!!! We will discuss 2 Ways to Add Vectors 1. Graphically (Today) 2. Component Addition (Later this week) Adding Vectors • When two vectors are added together, the results are a vector called a RESULTANT. • In order to find the resultant, you must draw the two vectors “head to tail.” Example: http://www.stmary.ws/highschool/physics/home/note s/forces/animations/AngleResultants.html Another Reason why vectors are important: http://www.stmary.ws/highschool/physics/home/notes/kine matics/MotionTerms/animations/boatsRiver.swf Find the Following Resultants 1. 5N + 5N = ? 3N = ? 2. + 6N Components • These two vectors are known as components. X - component • In physics, we often break vectors down into horizontal (x) and vertical (y) components. Y - component • Every vector can be separated into two equivalent vectors. Calculating Components • Use the Pythagorean Theorem to calculate the missing component for each velocity below. a2 + b2 = c2 5 m/s ? 8 m/s 4 m/s ? 8.9 m/s ? 6.24 m/s 3 m/s - 8 m/s 5 m/s 4 m/s a2 + b2 = c2 (-8)2 + 42 = c2 64 + 16 = c2 80 = c2 c = 8.9 m/s ? - 8 m/s 4 m/s Right Triangle Trig. Q: WHAT ARE TRIG FUNCTIONS? A: Sin Cos Tan • Three functions that are useful when dealing with right triangles. • These functions are used to determine missing sides of right triangles when other information is known. Sine (sin) • The Sin of any angle is equal to the ratio of the length of the “opposite side” and the length of the “hypotenuse.” O Sin (ø) = H H O ø Cosine (cos) • The Cosine of any angle is equal to the ratio of the length of the “adjacent side” and the length of the “hypotenuse.” A Cos (ø) = H H A ø Tangent (tan) • The Tangent of any angle is equal to the ratio of the length of the “opposite side” and the length of the “adjacent side.” O Tan (ø) = A O A ø An easy way to remember… • What should you do when you have dirty feet? »SOHCAHTOA (pronounced: Soak A Toe.. Uhhh….) SOH Sine Opposite Hypotenuse CAH Cosine Adjacent Hypotenuse TOA Tangent Opposite Adjacent What is a Function? • 2x = Y • What do I get for Y as I plug in numbers for x? Sin (x) Cos (x) and Tan (x) do the same thing!!!! x y Plug in a number for the angle: 0 0 (x degrees) 1 2 and you get back a different number (usually a decimal). 2 4 Next, following SOHCAHTOA, the number you get back from the function, is equal to O/H, A/H, or O/A! Trig. Practice Problems #1 a. Find x. b. Find y. c. Find q. d. Find r. q 9 25o r sin (30) = x/20 cos (30) = y/20 sin (25) = 9/q tan (25) = 9/r .5 = x/20 .866 = y/20 .423 = 9/q .466 = 9/r (.5)20 = x (.866)20 = y .423q = 9 .466r = 9 x = 10 y = 17.32 q = 21.3 r = 19.3 Today’s Objective By the end of today… 1.You must be able to find the magnitude and direction of a resultant mathematically. Weekly Homework • Read Ch. 3 Section 3 • Answer Chapter Review Questions On Pg. 109: 24, 25, 27, 29, 31, 32, 34 Trig. Problems #1 From a point on the ground 25 meters from the foot of a tree, the angle of elevation of the top of the tree is 32º. (Meaning you have to look up 32o to be looking at the top of the tree.) Find to the nearest meter, the height of the tree. HINT: Draw a diagram showing the information. m Trig. Practice #2 A ladder leans against a building. The foot of the ladder is 6 feet from the building. The ladder reaches height of 14 feet on the building. a. Find the length of the ladder to the nearest foot. Trig. Practice #2 b. What is the angle between the top of the ladder and the wall it leans on? Trig. Practice Problem #3 3. A plane travels 25 km at an angle 35 degrees to the ground and then changes direction and travels 12.5 km at an angle of 22 degrees to the ground. What is the magnitude and direction of the total displacement? Quiz! • The Vectors listed below represent the velocity of an airplane and the velocity of wind. Each Plane is trying to Go North. Draw a Vector Diagram and Calculate the magnitude and direction of the resulting velocity. 1. Wind: 50 m/s East Plane Engines: 200 m/s North 2. Wind 30 m/s East Plane Engines: 100 m/s North 3. Wind 120 m/s West Plane Engines: 175 m/s North Ch. 3.3 Projectile Motion Physics Mr. Pacton What is a projectile? • • A projectile is an object upon which the only force acting is gravity. There are a variety of examples of projectiles: 1. An object dropped from rest 2. An object which is thrown vertically or horizontally 3. An object which is thrown upwards at any angle NOTE: When we talk about projectiles we ignore the effects of air resistance and/or wind. The only force acting on a projectile must be gravity!!! Breaking it up… • It is extremely useful in physics to break up this motion into vertical and horizontal parts. Horizontal Motion Vertical Motion No Force Affecting Motion The force of gravity acts downward causing the projectile to “fall” No Acceleration Downward acceleration "g" is at ~ 10 m/s2 Constant Velocity Changing Velocity (increasing by ~10 m/s each second) Projectile Motion Equations Now lets look at an example! Vx = constant Vy = -gt increasing downward Vx *ONLY for horizontally launched projectile. Vy 1. So what is the horizontal velocity at 3 seconds for the cannon above? Vx = constant = 100 m/s 2. What is the vertical velocity at 3 seconds for the cannon above? Vy = -gt = -10x3 ~ -30 m/s 3. What is the total velocity at 3 seconds for the cannon above? (Hint: you have to add the two vectors!!!! A2+B2=C2) 2 2 2 2 2 V = Vx + Vy = 100 + 30 = 10,000 + 900 2 V = 10,900 Now take Square root to find V = 104.4 m/s Vocabulary: Range/Height height range The range of a projectile is how far it travels horizontally. The height is how high up into the air it goes. Projectile Motion Equations • We can also break down the displacement into Horizontal and Vertical parts! (x and y) Projectile Motion Equations • You know that Horizontally, the ball has a constant Velocity. So the total horizontal displacement, Dx, is: Dx = vixt • Vertically, the Velocity increases downward with an acceleration equal to g. The total vertical displacement, Dy, is: 2 Dy = viyt – ½gt Practice Problem • A cannon ball is fired horizontally off a cliff at a velocity of 100 m/s. What is the magnitude and direction of the velocity vector after 5 seconds? Projectile Motion Equations • Not every projectile is launched horizontally. • So the question remains, what general equations can we apply to any projectile? » Now you need to combine and use your knowledge of Trigonometry and Vector Math! Initial Velocity: V ø Vx Vix= V cos (ø) Vy Viy = V sin (ø) And to Find V at any time: 2 2 2 V = Vx + Vy Velocity at any time (t): Vx = Vix = V cos (ø) Vy = V sin (ø) - gt Mathematical Summary Breaking Down the Components Horizontal Vertical 0 g = -10 m/s 2 Velocity constant Vx = Vix changing Vy = Viy - gt V2 = Vx2 + Vy2 Displacement Dx = vixt Dy = viyt – ½gt2 D2= x2 + y2 Acceleration Sum And Vix = (Vi)(Cos q) Viy = (Vi)(Sin q) Quick Lab • Look at Pg. 97 in your book. • Complete the Quick Lab and answer all questions. (Turn in 1 paper per table) • Additionally, Answer the following: 1. What is g, the acceleration due to gravity, and does it affect both projectiles equally? 2. Describe what you have learned about the vertical motion of projectiles. (Minimum 3 sentences.) Practice Problem • A water balloon is launched with a speed of 40 m/s at an angle of 60 degrees to the horizontal. • Determine the horizontal and vertical components of the initial velocity. • Determine the time that the water balloon is in the air. • Determine the range of the water balloon.