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Learn how to analyze the vertical and horizontal components in 2D projectile motion scenarios using equations and real-life examples.
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Self evaluation • Tickets • Class officers
How are the vertical and horizontal components analyzed in 2 dimensional projectile motion? EQ: http://www.youtube.com/watch?v=0epHS1D-Swo http://www.youtube.com/watch?v=D9wQVIEdKh8
Projectile -An object upon which the only force acting is gravity which causes the object to curve downward along a parabolic path known as a trajectory
HORIZONTAL LAUNCH Vyi= 0 m/s not moving vertically yi> 0 m launched above ground Vxi> 0 m/s moving horizontally https://www.youtube.com/watch?v=EREQEy2tOxo ANGLE LAUNCH Vi > 0 m/s find x and y components to get Vxi and Vyi Highest point (ymax) Vy= 0 m/s Half of time http://www.youtube.com/watch?v=3wAjpMP5eyo 2 types of projectile:
Projectile Characteristics 2 types of projectile: YES - GRAVITY NO Yes – downward g = -9.80 m/s2 NO CHANGING CONSTANT
Check Point 1. Supposing a snowmobile is equipped with a flare launcher that is capable of launching a sphere vertically (relative to the snowmobile). If the snowmobile is in motion and launches the flare and maintains a constant horizontal velocity after the launch, then where will the flare land (neglect air resistance)? a. in front of the snowmobile b. behind the snowmobile c. in the snowmobile 2. Consider these diagrams in answering the following questions. Which diagram (if any) might represent ... a. ... the initial horizontal velocity? b. ... the initial vertical velocity? c. ... the horizontal acceleration? d. ... the vertical acceleration? e. ... the net force?
Remember the KINEMATIC equations??
Equations HORIZONTAL LAUNCH constant Things to note: *g= 9.80 m/s2 *viy = 0 m/s
Equations ANGLE LAUNCH constant
Example 1 of horizontal projectile • A movie director is shooting a scene that involves dropping a stunt dummy out of an airplane and into a swimming pool. The plane is 10.0 m above the ground, traveling at a velocity of 22.5 m/s in the positive x direction. The director wants to know where in the plane’s path the dummy should be dropped so that it will land in the pool. What is the dummy’s horizontal displacement?
G: Initial velocity is zero Because it is in free fall, a becomes g Usually if a vertical equation is used to solve for time, then a horizontal equation can be used to find the unknown quantity or vice versa U: Velocity initial in x direction E: First solve for t with given info S: S: Vx = 5.56 m/s
Example 3 of Horizontal Projectile Florence Griffith-Joyner of the United States set the women’s world record for the 200 m run by running with an average speed of 9.37 m/s. Suppose Griffith-Joyner wants to jump over a river. She runs horizontally from the river’s higher bank at 9.37 m/s and lands on the edge of the opposite bank. If the difference in height between the two banks is 2.00 m, how wide is the river?
G: Initial velocity is zero Because it is in free fall, a becomes g Usually if a vertical equation is used to solve for time, then a horizontal equation can be used to find the unknown quantity or vice versa U: X – width of river E: First solve for t with given info S: S: X = 5.99 m
Example 1 of Projectiles at an Angle A baseball is thrown at an angle of 25°relative to the ground at a speed of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long was it in the air? How high was the tallest spot in the ball’s path?
Example 2 of Projectiles at an Angle A soccer player kicks a soccer ball towards the goal. If he kicks it with a velocity of 24 m/s at an angle of 31 degrees, how far will the ball travel?
G = 9.80 m/s2 Vyf = 0 m/s Vi = 24 m/s = 31° G: U: Distance in horizontal (x) direction E: S: S: X = 52 m
A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the horizontal, Find each of the following. Assume that forces from the air on the ball are negligible. • the ball’s hang time b. the ball’s maximum height c. the horizontal distance the ball travels before hitting the ground Example 3 of Projectiles at an Angle
G = 9.80 m/s2 Vyf = 0 m/s Vi = 27.0 m/s = 30.0° G: U: Time in air (t) Max height (y) Distance in horizontal direction (x) E: a. S: t= 1.377*2 = 2.755 S: T = 2.76 s
G = 9.80 m/s2 Vyf = 0 m/s Vi = 27.0 m/s = 30.0° G: Maximum hieght occurs at half the hang time. T= 2.76/2 = 1.38 s. Thus, b. E: S: S: Y = 9.30 m
E: Distance in x direction S: S: