1. Sect. 2-7: Falling Objects

2. Freely Falling Objects
One of the most important & common special cases of motion with constant acceleration:
“FREE FALL”
Objects falling due to gravity near the surface
of Earth. Neglect air resistance. Near Earth’s
surface, all objects experience approximately
the same acceleration due to gravity.
Use the one-dimensional constant
acceleration equations
(with some changes in notation,
as we will see)

3. Experiment:
Ball & light piece of paper dropped at the same time. Repeated with wadded up paper.
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. Rock & feather dropped at the same time in air. Repeated in vacuum.
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/s2. At a given location on the Earth and in the absence of air resistance, all objects fall with the same constant acceleration.

5. Experiment finds that the acceleration of falling objects (neglecting air resistance) is always (approximately) the same, no matter how light or heavy the object.
Acceleration due to gravity, a  g
g = 9.8 m/s2 (approximately!)

6. Acceleration due to gravity, g = 9.8 m/s2
Acceleration of falling objects is always the same, no matter how light or heavy.
Acceleration due to gravity, g = 9.8 m/s2
First proven by Galileo Galilei
Legend: Dropped
objects off of the
leaning tower of Pisa.

7. A COMMON MISCONCEPTION!

8. Acceleration due to gravity g = 9.8 m/s2 (approximately)
Depends on location on Earth, latitude, & altitude:

9. Note: My treatment is slightly different than the book’s, but it is, of course, equivalent!
To treat motion of falling objects, use the same equations we already have, but change notation slightly:
Replace a by g = 9.8 m/s2
But in the equations it could have a + or a - sign in front of it! Discuss this next!
Usually, we consider vertical motion to be in the y direction, so replace x by y and x0 by y0 (often y0 = 0)

10. NOTE!!!
Whenever I (or the author!) write the symbol g, it ALWAYS means the POSITIVE numerical value 9.8 m/s2! It NEVER is negative!!! The sign (+ or -) of the gravitational acceleration is taken into account in the equations we now discuss!

11. Sign of g in 1d Equations Magnitude (size) of g = 9.8 m/s2 (POSITIVE!)
But, acceleration is a vector (1 dimen), with 2 possible directions.
Call these + and -.
However, which way is + and which way is - is ARBITRARY & UP TO US!
May seem “natural” for “up” to be + y and “down” to be - y, but we could also choose (we sometimes will!) “down” to be + y and “up” to be - y
So, in equations g could have a + or a - sign in front of it, depending on our choice!

12. Directions of Velocity & Acceleration
Objects in free fall ALWAYS have downward acceleration.
Still use the same equations for objects thrown upward with some initial velocity v0
An object goes up until it stops at some point & then it falls back down. Acceleration 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!

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

14. Equations for Bodies in Free Fall
Written taking “up” as + y!
v = v0 - g t (1)
y = y0 + v0 t – (½)gt2 (2)
v2 = (v0)2 - 2g (y - y0) (3)
v = (½)(v + v0) (4)
g = 9.8 m/s2
Usually y0 = 0. Sometimes v0 = 0

15. Equations for Bodies in Free Fall
Written taking “down” as + y!
v = v0 + g t (1)
y = y0 + v0 t + (½)gt2 (2)
v2 = (v0)2 + 2g (y - y0) (3)
v = (½)(v + v0) (4)
g = 9.8 m/s2
Usually y0 = 0. Sometimes v0 = 0

16. Example 2-10: Falling from a Tower
A ball is dropped (v0 = 0) from a tower m high. How far will it have fallen after time t1 = 1 s, t2 = 2 s, t3 = 3 s?
Note: y is positive
DOWNWARD!
v = gt
y = (½) gt2
a = g = 9.8 m/s2
v1 = (9.8)(1)
= 9.8 m/s
v2 = (9.8)(2)
= 19.6 m/s
v3 = (9.8)(3)
= 29.4 m/s

17. Example 2-11: Thrown Down From a Tower
A ball is thrown downward with an initial velocity of v0 = 3 m/s, instead of being dropped. What are it’s position & speed after t1 = 1 s & t2 = 2 s? Compare with the dropped ball.
Note: y is positive
DOWNWARD!
v = v0 + gt
y = v0t + (½)gt2
a = g = 9.8 m/s2
y1, v1
y2, v2
y3, v3

18. Note: y is positive UPWARD!
v = 0 here, but a = - g!
Examples 2-12, 2-14, 2-15
A person throws a ball upward into the air with an initial velocity of v0 = 15.0 m/s. Calculate
Time to top
= ½ round
trip time
a. The time to reach the maximum height.
b. The maximum height.
c. The time to come back to the hand.
d. The velocity when it returns to the hand.
Note: y is positive UPWARD!
v = v0 – gt, y = v0t - (½)gt2
v2 = (v0)2 - 2g(y - y0)
 vC = - v0
= -15 m/s
v0 = 15 m/s 

19. Example: Not a bad throw for a rookie!
Problem: A stone is thrown at point (A) from the top of a building with an initial velocity of v0 = 19.2 m/s straight upward. The building is H = 49.8 m high, and the stone just misses the edge of the roof on its way down, as in the figure. Answer these questions:
a) Calculate the time at which the stone reaches its maximum height.
b) Calculate the maximum height of the stone above the rooftop.
c) Calculate the time at which the stone returns to the level of the thrower
d) Calculate the velocity of the stone at this instant.
e) Calculate the velocity & position of the stone
at time t = 5 s

20. Example: Ball Thrown Up at the Edge of a Cliff
A ball is thrown up at speed 15.0 m/s by a person on the edge of a cliff. The ball can fall to the base of the cliff 50.0 m below. Ignore air resistance. Calculate:
a. The time it takes the ball to reach the base of the cliff.
b. The total distance traveled by the ball.
Figure Caption: Example 2–20. The person in Fig. 2–30 stands on the edge of a cliff. The ball falls to the base of the cliff, 50.0 m below.
Solution: a. We use the same quadratic formula as before, we find t = 5.07 s (the negative solution is physically meaningless).
b. The ball goes up 11.5 m, then down 11.5 m + 50 m, for a total distance of 73.0 m.