Motion in Two Dimensions Chapter 7.2
Homework P. 158 P.160. Problems 1-9 (handout) 9,10,11 P.160. 12,13. Problems 1-9 (handout) Due Thursday PM P. 172 39, 40, 44, 45, 46 Due Monday
Projectile Motion What is the path of a projectile as it moves through the air? Parabolic? Straight up and down? Yes, both are possible. What forces act on projectiles? Only gravity, which acts only in the negative y-direction. Air resistance is ignored in projectile motion.
Choosing Coordinates & Strategy For projectile motion: Choose the y-axis for vertical motion where gravity is a factor. Choose the x-axis for horizontal motion. Since there are no forces acting in this direction (of course we will neglect friction due to air resistance), the speed will be constant (a = 0). Analyze motion along the y-axis separate from the x-axis. If you solve for time in one direction, you automatically solve for time in the other direction.
The Trajectory of a Projectile What does the free-body diagram look like for force? Fg
The Vectors of Projectile Motion What vectors exist in projectile motion? Velocity in both the x and y directions. Acceleration in the y direction only. vx (constant) ax = 0 vy (Increasing) ay Trajectory or Path Why does the velocity increase in the y-direction? Gravity. Why is the velocity constant in the x-direction? No force acting on it.
Ex. 1: Launching a Projectile Horizontally A cannonball is shot horizontally off a cliff with an initial velocity of 30 m/s. If the height of the cliff is 50 m: How far from the base of the cliff does the cannonball hit the ground? With what speed does the cannonball hit the ground?
Diagram the problem vi 50m a = -g Fg = Fnet vf = ? vx vy x = ?
State the Known & Unknown xi = 0 vix = 30 m/s yi = 0 viy = 0 m/s a = -g y = -50 m Unknown: x at y = -50 m vf = ?
Perform Calculations (y) y-direction: vy = -gt y = viyt – ½ gt2 Using the first formula above: vy = (-9.8 m/s2)(3.2 s) = 31 m/s
Perform Calculations (x) x-Direction x = vixt x = (30 m/s)(3.2 s) = 96 m from the base. Using the Pythagorean Theorem: v = vx2 + vy2 v = (30 m/s)2 + (31 m/s)2 = 43 m/s
Ex. 2: Projectile Motion above the Horizontal A ball is thrown from the top of the Science Wing with a velocity of 15 m/s at an angle of 50 degrees above the horizontal. What are the x and y components of the initial velocity? What is the ball’s maximum height? If the height of the Science wing is 12 m, where will the ball land? P 5/6, 7/8 wed 10/22
Diagram the problem Fg = Fnet a = -g vi = 15 m/s viy vix y = 50° Ground x = ?
State the Known & Unknown xi = 0 yi = 12 m vi = 15 m/s = 50° a = -g Unknown: ymax = ? t = ? x = ? viy = ? vix = ?
Perform the Calculations (ymax) y-direction: Initial velocity: viy = visin viy = (15 m/s)(sin 50°) viy = 11.5 m/s Time when vfy = 0 m/s: vfy = viy – gt t = viy / g t = (11.5 m/s)/(9.81 m/s2) t = 1.17 s Determine the maximum height: ymax = yi +viyt – ½ gt2 ymax = 12 m + (11.5 m/s)(1.17 s) – ½ (9.81 m/s2)(1.17 s)2 ymax = 18.7 m vi = 15 m/s vxi vyi = 50°
Perform the Calculations (t) Since the ball will accelerate due to gravity over the distance it is falling back to the ground, the time for this segment can be determined as follows Time when ball hits the ground: ymax = viyt – ½ gt2 Since yi can be set to zero as can viy, t = 2*ymax/g t = 2(18.7 m)/ (9.81 m/s2) t = 1.95 s By adding the time it takes the ball to reach its maximum height to the time it it takes to reach the ground will give you the total time. ttotal = 1.17 s + 1.95 s = 3.12 s
Perform the Calculations (x) x-direction: Initial velocity: vix = vicos vix = (15 m/s)(cos 50°) vix = 9.64 m/s Determine the total distance: x = vixt x = (9.64 m/s)(3.12 s) x = 30.1 m vi = 15 m/s vxi vyi = 50°
Analyzing Motion in the x and y directions independently. x-direction: dx = vix t = vfxt vix = vicos y-direction: dy = ½ (vi + vf) t = vavg t vf = viy + gt dy = viy t + ½ g(t)2 vfy2 = viy2 + 2gd viy = visin
Key Ideas Projectile Motion: Gravity is the only force acting on a projectile. Choose a coordinate axis that where the x-direction is along the horizontal and the y-direction is vertical. Solve the x and y components separately. If time is found for one dimension, it is also known for the other dimension.
There are only so many types of problems: Vix = constant Viy = 0 Viy >0 (shooting upwards at an angle) Viy < 0 (shotting down at an angle) a = g a <>g