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This resource explores the physics of projectiles launched at an angle, focusing on motion in two dimensions. It covers key concepts including constant velocity on the x-axis, constant acceleration on the y-axis, and parabolic trajectories. Learn how to calculate initial velocities, maximum heights, and ranges for various projectile scenarios. The examples provided illustrate the principles at play, including complementary angles and symmetry in projectile motion. Ideal for students and educators looking to deepen their understanding of physics in action.
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Projectiles at an Angle Physics Mrs. Coyle
Projectiles at an Angle http://www.physicsclassroom.com
Projectiles at an Angle • Motion in two dimensions • x-axis constant velocity • y-axis constant acceleration • Parabolic trajectory
Initial Velocity • vxi=vcosq • vyi=vsinq v
Equations for Projectiles at an Angle • Initial Velocity is at an angle with the horizontal • vx=constant • vy=vyi+at a=-10m/s2 • Dy=vyit+ 1at2 2
Additional Key Points • At the max height: vy=0 vx=constant=vxi • Maximum range when q=45 degrees in the absence of air resistance. • Complementary angles result in the same range. • Symmetry • Resultant velocity is tangent to the curve at the point of study.
Demonstration Link Projectiles at an Angle
Example 1 An arrow is shot with an initial x velocity of 5m/s and an initial y velocity is 15m/s. • How much time will pass for the projectile to reach max height? • What is the max height? • What is the range? Answer: a) t=1.5s , b)y= 11.25m, c) x=15m
Example 2 A cannonball is shot with an initial x velocity of 60m/s and an initial y velocity of 25m/s. a)What is the max height? b)What is the range?Answer: a) t=2.5s, y=31.25m b) x=300m
Example 3 Using vx=vcosq, vy=vsinq A cannonball is shot at 60 degrees with an initial velocity of 40m/s. a)What is the max height? b)What is the range? c)At what angle should the projectile be shot to achieve max range and what is the max range? a)Vx=20m/s, Vy=34.64m/s, t max=3.5s, y=61mb) 140m, c) 45o , 163m