1. Freely Falling Objects

2. Important & common special case of uniformly accelerated motion: “FREE FALL” Objects falling in Earth’s gravity. Neglect air resistance. Use one dimensional uniform acceleration equations (with some changes in notation, as we will see) Freely Falling Objects

3. Near the surface of the Earth, all objects experience approximately the same acceleration due to gravity. This is one of the most common examples of motion with constant acceleration. In the absence of air resistance, all objects fall with the same acceleration, although this may be tricky to tell by testing in an environment where there is air resistance.

4. Falling Objects Experiment: –Ball & light piece of paper dropped at the same time. Repeated with wadded up paper.

5. Experiment: –Rock & feather dropped at the same time in air. Repeated in vacuum. The acceleration due to gravity at the Earth’s surface is Approximately 9.80 m/s 2. At a given location on the Earth & in the absence of air resistance, all objects fall with the same constant acceleration.

6. Experiment finds that the acceleration of falling objects (neglecting air resistance) is always (approximately) the same, no matter how light or heavy the object. The magnitude of the acceleration due to gravity, a  g g = 9.8 m/s 2 (approximately!)

7. The acceleration of falling objects is always the same, no matter how light or heavy. Acceleration due to gravity, g = 9.8 m/s 2 First proven by Galileo Galilei A Legend: He dropped objects off of the leaning tower of Pisa.

8. The magnitude of the acceleration due to gravity: g = 9.8 m/s 2 (approximately!) It has a slight dependence on the location on Earth, on the latitude & the altitude:

9. A Common Misconception!

10. The Algebraic Sign (+ or - ?) in the Kinematics Equations of the One-Dimensional Vector Gravitational Acceleration Note: My treatment is slightly different than the book’s, but it is equivalent! To treat motion of falling objects, we use the same equations we already have, but change notation slightly: Replace a by g = 9.8 m BUT in the equations, it could have a + or a - sign in front of it! We discuss this next! Usually, we consider vertical motion to be in the y direction & so replace x by y and x 0 by y 0 (often y 0 = 0)

11. NOTE!!! Whenever I (or the author!) write the symbol g, it ALWAYS means the POSITIVE numerical value 9.8 m/s 2 !! It is NEVER negative!!! The sign (+ or -) of the one-dimensional gravitational acceleration VECTOR is taken into account in the Equations we now discuss!

12. The Sign of g in the 1d Equations The magnitude (size) of g = 9.8 m/s 2 (POSITIVE!) –But, acceleration is a (1 dimensional) VECTOR with 2 possible directions. –Call these + and -. –However, which way is + and which way is – is ARBITRARY & UP TO US! –It may seem “natural” for “up” to be + y & “down” to be - y, but we could also choose (we sometimes will!) “down” to be + y and “up” to be - y So, in the equations g could have a + or a - sign in front of it, depending on our choice!

13. Directions of Velocity & Acceleration Objects in free fall ALWAYS have DOWNWARD acceleration. We still use the same equations for objects thrown upward with some initial velocity v 0 An object goes up until it stops at some point & then it falls back down. The acceleration vector is always g in the downward direction. For the first half of flight, the velocity is UPWARD.  For the first part of the flight, velocity & acceleration are in opposite directions!

14.  VELOCITY & ACCELERATION ARE NOT NECESSARILY IN THE SAME DIRECTION!

15. Equations for Objects in Free Fall Written taking “up” as + y v = v 0 - gt (1) y = y 0 + v 0 t - (½)gt 2 (2) (v) 2 = (v 0 ) 2 - 2g(y - y 0 ) (3) v avg = (½)(v + v 0 ) (4) g = 9.8 m/s 2 Often, y 0 = 0. Sometimes v 0 = 0

16. Equations for Objects in Free Fall Written taking “down” as + y v = v 0 + gt (1) y = y 0 + v 0 t + (½)gt 2 (2) (v) 2 = (v 0 ) 2 + 2g(y - y 0 ) (3) v avg = (½)(v + v 0 ) (4) g = 9.8 m/s 2 Often, y 0 = 0. Sometimes v 0 = 0

17. Free Fall Examples

18. Example Falling from a tower (v 0 = 0) Note! Take y as positive DOWNWARD! v = at y = (½)at 2 a = g = 9.8 m/s 2

19. Example Falling from a tower (v 0 = 0) Note! Take y as positive DOWNWARD! v = at y = (½)at 2 a = g = 9.8 m/s 2

20. Example Falling from a tower (v 0 = 0) Note! Take y as positive DOWNWARD! v = at y = (½)at 2 a = g = 9.8 m/s 2

21. Example Falling from a tower (v 0 = 0) v 1 = (9.8)(1) = 9.8 m/s v 2 = (9.8)(2) = 19.6 m/s v 3 = (9.8)(3) = 29.4 m/s Note! Take y as positive DOWNWARD! v = at y = (½)at 2 a = g = 9.8 m/s 2

22. Example: Thrown Down From a Tower A ball is thrown downward with initial velocity v 0 = 3.0 m/s, instead of being dropped. (a) Position after t = 1.0 s & 2.0 s? y = v 0 t + (½)at 2 t = 1.0 s; y = (3)(1) + (½)(9.8)(1) 2 = 7.9m t = 2.0s ; y = (3)(2) + (½)(9.8)(2) 2 = 25.6 m Photo of the leaning tower of Pisa

23. Example: Thrown Down From a Tower A ball is thrown downward with initial velocity v 0 = 3.0 m/s, instead of being dropped. (a) Position after t = 1.0 s & 2.0 s? y = v 0 t + (½)at 2 t = 1.0 s; y = (3)(1) + (½)(9.8)(1) 2 = 7.9m t = 2.0s ; y = (3)(2) + (½)(9.8)(2) 2 = 25.6 m Photo of the leaning tower of Pisa

24. Example: Thrown Down From a Tower A ball is thrown downward with initial velocity v 0 = 3.0 m/s, instead of being dropped. (a) Position after t = 1.0 s & 2.0 s? y = v 0 t + (½)at 2 t = 1.0 s; y = (3)(1) + (½)(9.8)(1) 2 = 7.9m t = 2.0s ; y = (3)(2) + (½)(9.8)(2) 2 = 25.6 m (b) Speed after t = 1.0 s & 2.0 s? v = v 0 + at t = 1.0 s; v = 3 + (9.8)(1) = 12.8m/s t = 2.0s ; v = 3 + (9.8)(2) = 22.6m/s Compare with speeds of a dropped ball. Photo of the leaning tower of Pisa

25. A Useful, Detailed Example v A = v 0 = 15 m/s choose y as positive upward  a = -g = - 9.8 m/s 2 Time to the top = (½) round trip time! v = 0 here but still have a = -g v C = -v A (= -v 0 ) A person throws a ball up into the air with initial velocity v 0 = 15.0 m/s. Questions: a. Time to the top? b. Round trip time? c. Maximum height? d. Velocity when it comes back to the start? e. Times when the height y = 8.0 m? Work in general on the white board!

26. Example: Ball thrown upward; the quadratic formula. For a ball thrown upward at an initial speed of v 0 = 15.0 m/s, calculate the times t the ball passes a point y = 8.0 m above the person’s hand.

27. Example: A ball thrown upward at the edge of a cliff. A ball is thrown up with initial velocity v 0 = 15.0 m/s, by a person standing on the edge of a cliff, so that it can fall to the base of the cliff 50.0 m below. Calculate: a. The time it takes the ball to reach the base of the cliff. b. The total distance traveled by the ball.

28. A stone is thrown at point (A) from the top of a building with initial velocity v 0 = 19.2 m/s straight up. The building is H = 49.8 m high, & the stone just misses the edge of the roof on its way down, as in the figure. Calculate: a) The time at which it reaches its maximum height. b) It’s maximum height above the rooftop. c) The time at which it returns to the thrower’s hand. d) It’s velocity when it returns to the thrower’s hand. e) It’s velocity & position at time t = 5 s. Example: Not a bad throw for a rookie!