Chapter 5 Notes PDF

Chapter 5---Projectile Motion Faucher  Vector and Scalar Quantities Scalar Scalar is just a number --- stand on a scale, it just tells you your weight Vector Vector is a number with a direction.  Projectile Motion Vector Addition...

Chapter 5---Projectile Motion Faucher  Vector and Scalar Quantities Scalar Scalar is just a number --- stand on a scale, it just tells you your weight Vector Vector is a number with a direction.  Projectile Motion Vector Addition + – A arrow is used to represent the magnitude and direction of a vector A velocity is sometimes a result of the addition or subtraction of vector quantities  Projectile Motion Components of Vectors Any single vector can be resolved into two components Component vectors form right angles to each other – a VERTICAL component of the velocity – a HORIZONTAL component of velocity  Projectile Motion Any projectile has both vertical and horizontal components. This path a projectile follows is a parabola.  Projectile Motion – Horizontal component of velocity stays constant – Vertical component of velocity has gravity acting on it --- so it changes in it’s path.  Projectile Motion Ideal path and actual path of a projectile are different – We look at ideal path — NO air resistance – If ideal --- then time up equals time down & velocity up equals velocity down  ADDING Vectors Mathematically 1. Sum up all the vectors directed in the x-direction (∑x). 2. Sum up all the vectors in the y-direction (∑y). 3. Redraw ∑x and ∑y head to tail. Include directions. 4. Draw Resultant Vector (from start to ending point) 5. Apply Pythagorean Theorem for size of Resultant The Pythagorean Theorem The sum of the areas of the two squares on the legs (A and B) equals the area of the square on the hypotenuse (C). C (hypotenuse) B (leg) 90º A (leg) Pythagoras of Samos C² = A² + B² (580 BC-490 BC)  Projectile Motion A cannonball shot from a cannon, a stone thrown in the air, a spacecraft circling Earth – all of these are examples of projectiles. Projectiles follow a curved path which may seem complicated, but if we look at the horizontal and vertical components of motion separately, these paths are somewhat simple.  Rules for Projectiles: All objects are pulled downward by gravity with a constant acceleration. g = -9.81 m/s2 The acceleration is the same whether an object is dropped or shot forward.  In the x direction, gravity has no effect. The velocity is constant, so the horizontal position of the object can be found by x = vt  In the y direction, gravity pulls downward just as if the object were dropped. The y position is given by y = ½ at2 + tvi But since the initial velocity is zero, we can change this to y = ½ at2 What is “a” during free fall? The actual path of the projectile is the resultant of the two vectors.  No matter the angle at which a projectile is launched, the vertical distance of fall beneath the idealized straight-line path (dashed straight lines) is the same for equal times. With no gravity the projectile would follow the straight-line path (dashed line). But because of gravity it falls beneath this line the same vertical distance it would fall if it were released from rest.  Equations HORIZONTAL (x) VERTICAL (y) vfx = dx/t vfy = viy + ay t dx = vix t + ½ *ax*t2 dy = viy t + ½ *ay*t2 vfx2 = vix2 + 2*ax x vfy2 = viy2 + 2*ay y How long is the ball in the air? X Y x= y= Let’s start by vx= viy= plugging in the ax= ay= numbers we know. vfy= t= How long is the ball in the air? X Y There is never acceleration in x=20m y=-5m the x direction, so it will vx= viy= 0m/s always be 0. 2 Gravity only affects the object ax = 0m/s ay=-9.8m/s2 in the y direction. vfy= Now take a few minutes to t=? solve. How long is the ball in the air? X Y x=20m y=-5m y = ½ at2 vx= viy= 0m/s -5 = ½ (-9.8) t2 ax= 0m/s2 ay=-9.8m/s2 -5 = -4.9t2 vfy= 1 = t2 t= 1 sec 1=t How fast is the ball thrown? X Y This question basically wants x=20m y=-5m you to find the velocity in the x direction. vx=? viy=0m/s 2 2 Look at the horizontal motion ax =0m/s a y =-9.8m/s section on your equation charts vfy= for hints to solve. t= 1 sec Now take a minute to solve. How fast is the ball thrown? X Y x=20m y=-5m Vx = x/t vx=20m/s viy=0m/s Vx =20m/1sec ax=0m/s2 ay=-9.8m/s2 Vx = 20m/s vfy= t= 1 sec What is the resultant velocity of the ball as it strikes the ground? X Y x=20m y=-5m vx=20m/s viy=0m/s ax=0m/s2 ay=-9.8m/s2 vfy= t= 1 sec What is the resultant velocity of X Y the ball as it strikes the ground? x=20m y=-5m 1st – Find the Final Velocity in vx=20m/s viy=0m/s the y direction ax=0m/s2 ay=-9.8m/s2 2nd – Use pythagorean theorem vfy= to find the resultant velocity t= 1 sec Take a few minutes to solve for the first part. What is the resultant velocity of the ball as it strikes the X Y ground? x=20m y=-5m Vfy = Viy + at vx=20m/s viy=0m/s Vfy = 0 + (-9.8)(1) ax=0m/s2 ay=-9.8m/s2 Vfy = -10 m/s vfy=-10m/s t= 1 sec vx=20m/s vfy=-10m/s What is the resultant velocity of the ball as it strikes the X Y ground? x=20m y=-5m Notice that you can move the Vfy over to make a right triangle. vx=20m/s viy=0m/s This allows us to use the ax=0m/s2 ay=-9.8m/s2 pythagorean theorem to find the impact or resultant velocity. vfy=-10m/s Now take a minute to solve for the t= 1 sec resultant. vx=20m/s vfy= -10 m/s ? What is the resultant velocity of the ball as it strikes the X Y ground? x=20m y=-5m a2 + b 2 = r 2 vx=20m/s viy=0m/s (20)2 + (-10)2 = r2 ax=0m/s2 ay=-9.8m/s2 400 + 100 = r2 vfy=-10m/s 500 = r2 t= 1 sec r = 22.4 m/s  A projectile traveling fast enough to fall around the earth rather than into it. (8km/s) Launch Speed less than 8000 m/s Launch Speed equal to 8000 m/s Projectile falls to Earth Projectile orbits Earth - Circular Path Launch Speed greater than 8000 m/s Projectile orbits Earth - Elliptical Path NASA  Examples of Projectiles Monkey Ballistics Juggling Moon feather