Projectile Motion 1 Preflight Responses A flatbed railroad

Projectile Motion 1

Preflight Responses • A flatbed railroad car is moving along a track at constant velocity. A passenger riding at the center of the train throws a ball straight up. Neglecting air resistance, where will the ball land? • • • Forward of the center of the train. tly! c e r r o c d e % answer 74 Right at the center of the train. Backward of the center of the train.

“x” and “y” components of motion are independent. l A man on a train tosses a ball straight up in the air. çView this from two reference frames: Reference frame on the moving train. Reference frame on the ground.

Frames of Reference

Projectile: An Object Moving Solely Under Influence of Gravity

Preflight Responses You and a friend are standing on level ground, each holding identical baseballs. At exactly the same time, and from the same height, you drop your baseball without throwing it while your friend throws her baseball horizontally as hard as she can. Which ball hits the ground first? ! Your ball y l t c re r o c d e r e w s n a Your friends ball 79% They both hit the ground at the same time.

Horizontally Launched Projectile

Projectile Motion Without gravity, the cannonball would follow a straight-line path as shown in the picture. With gravity, the cannonball falls the same distance below the gravity-free path as a cannonball dropped from the same height.

Motion in Freefall Horizontal Motion Acceleration NO Velocity Constant Vertical Motion

Motion in Freefall Horizontal Motion Acceleration NO Velocity Constant Vertical Motion YES – “g” is downward at 9. 8 m/s/s Changing by 9. 8 m/s, downward each second

Independence of x & y Motions For Vx= 5 m/s t (s) Vy(m/s) Y (m) X (m) 0 0 1 5 2 10 3 15 4 20 t v xt

Independence of x & y Motions For Vx= 5 m/s Y (m) X (m) t (s) Vy(m/s) 0 0 0 1 9. 8 5 2 19. 6 10 3 29. 4 15 4 39. 2 20 t 9. 8 t v xt

Independence of x & y Motions For Vx= 5 m/s t (s) Vy(m/s) Y (m) X (m) 0 0 1 9. 8 4. 9 5 2 19. 6 10 3 29. 4 44. 1 15 4 39. 2 78. 4 20 t 10 t 4. 9 t 2 vxt

Problem: • The pilot of a hovering helicopter drops a lead brick from a height of 1000 m. How long does it take to reach the ground and how fast is it moving when it gets there? (neglect air resistance) 1000 m

Problem: • First choose coordinate system. -Origin and y-direction • Next write down position equation: y 1000 m Realize that y = 1000 m

Problem: • Solve for time t when y = 1000 m • y • To find the velocity: y 0 = 1000 m • Evaluate vy: y = 1000 m

Problem 2: • A helicopter is traveling toward the west at 30 m/s. The pilot of drops a lead brick from a height of 1000 m. How long does it take to reach the ground and what is its velocity when it gets there? How far west of the release point does the brick hit the ground? (neglect air resistance) 1000 m

Problem 2: Since the time to reach the ground is INDEPENDENT OF HORIZONTAL MOTION x Find the time as before – using the : vertical motion equation: y 1000 m Find the horizontal position (x) using the constant velocity equation for the horizontal motion equation: x

Problem 2: To find the speed when it reaches the ground, we must use vector addition to combine the x and y velocities. x The y-velocity, from before is: y The x-velocity is constant 1000 m θ The magnitude of the velocity is: v The angle the velocity makes with the horizontal is: x vy = 140 m/s