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Learn the principles of projectile motion with real-life examples and clear solutions, utilizing kinematics equations for both horizontal and vertical components. Understand trajectories and velocities in launched horizontally and at an angle scenarios.
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PROJECTILE MOTION 2-D MOTION
Projectile Motion • Curve path or trajectory • x and y motion at the same time • Use the 4 kinematics equations • x direction: x, v, t • y direction: y, vf, vo, a, t • 2 types • Launched horizontally • Launched at an angle
Launched Horizontally • x-direction • NO a • vº = vf • y-direction • Δ uniformly • a = g • vº = 0 m/s • ty = tx
Example 1 • A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high. • How long does it take the stone to reach the bottom of the cliff? • How far from the base of the cliff does the stone strike the ground? • What are the horizontal and vertical components of the velocity of the stone just before it hits the ground?
Example 1 • Draw Diagram • Make a table
Example 1 • Solve for time • y = vot + ½gt2 = ½gt2 • 2y 2(78.4 m) t = -------- = -------------- √ g √ 9.8 m/s/s = 4.0 s
Example 1 • Solve for x • x = vxt = (5.0 m/s) (4.0 s) = 20. m
Example 1 • Solve for vx and vy final • vx is constant = 5.0 m/s • vy is changing • vy = vº+ gt = gt = (9.8m/s/s) (4.0 s) = 39 m/s
Example 2 • A steel ball rolls with constant velocity across a tabletop 0.950 m high. It rolls off and hits the ground 0.352 m horizontally from the edge of the table. How fast was the ball rolling?
Example 2 • Draw Diagram • Make a table
Example 2 • Solve for time • y = vot + ½gt2 = ½gt2 • 2y 2(0.950 m) t = -------- = -------------- √ g √ 9.8 m/s/s = 0.440 s
Example 2 • Solve for vx • x vx = ------ t • 0.352 m vx = ------------- 0.440 s = 0.800 m/s
Example 3 • An auto, moving too fast on a horizontal stretch of mountain road, slides off the road, falling into deep snow 43.9 m below the road and 87.7 m beyond the edge of the road. • How long did the auto take to fall? • How fast was it going when it left the road? • What was its acceleration 10 m below the edge of the road?
Example 3 • Draw Diagram • Make a table
Example 3 • Solve for time • y = vot + ½gt2 = ½gt2 • 2y 2(43.9 m) t = -------- = -------------- √ g √ 9.8 m/s/s = 2.99 s
Example 3 • Solve for vx • x vx = ------ t • 87.7 m vx = ------------- 2.99 s = 29.3 m/s
Example 3 Acceleration is only due to gravity.
Launched at an Angle • x-direction • NO a • vº constant • y-direction • Δ uniformly • a = g • tup = tdown • ty = tx • Solve x and y
Example 4 • A player kicks a football from ground level with a velocity of magnitude of 27.0 m/s at an angle of 30.0º above the horizontal. • Find the hang time. • Find the distance the ball travels before it hits the ground. • Find its maximum height.
Example 4 • Draw Diagram • Make a table
Launched at Angle • Solve for initial vx and vy • vx = v cosθ = (27 m/s) cos 30º = 23 m/s • vy = v sinθ = (27 m/s) sin 30º = 14 m/s
Example 4 • Solve for time • vy = vo + gt - 14 m/s = 14 m/s + (-9.8 m/s2) t -28 m/s = -9.8 m/s2 t t = 2.8s
Example 4 • Solve for time • y = vot + ½gt2 = ½gt2 • 2y 2(43.9 m) t = -------- = -------------- √ g √ 9.8 m/s/s = 2.99 s
Example 4 • Solve for time • y = vot + ½gt2 = ½gt2 • 2y 2(43.9 m) t = -------- = -------------- √ g √ 9.8 m/s/s = 2.99 s
Example 5 • A rude tourist throws a peach pit horizontally with a 7.0 m/s velocity out of an elevator cage. • If the elevator is not moving, how long will the pit take to reach the ground, 17.0 m below? • How far from the elevator will the pit land? • He throws the next pit when the elevator is at the same height but moving upward at a constant 8.5 m/s velocity. How long will it take this pit to land? • How far away will this pit land?
Example 5 • Draw Diagram • Make a table
Graphing Projectile Motion Height Horizontal Distance
Graphing Projectile Motion Height Time
Graphing Projectile Motion Vertical Speed Time
Graphing Projectile Motion Horizontal Velocity Time
Graphing Projectile Motion Horizontal Distance Time