1. Free Fall

2. Galileo’s Discovery
Which falls with more acceleration, your physics text book or a piece of paper?
The text book will fall faster.
What happens if the piece of paper sits on top of the text book?
Both will fall equally the same. Why?
What happens if both objects are in a vacuum?

3. Galileo’s Discovery
Galileo noticed that objects will fall at the same rate (as long as air resistance does not affect said object)
Free Fall – is the motion of an object when gravity is the only significant force acting on it.
Free Fall Acceleration – is the acceleration of an object due only to the effect of gravity, regardless of its mass.
Free Fall Acceleration on earth is 9.8 m/s2 down or m/s2.

4. Acceleration Due to Gravity
Acceleration Due to Gravity Comparison
Body
Mass [kg]
Radius [m]
Acceleration Due  to Gravity, "g" [m/s²]
g / g-Earth
Sun
1.99 x 1030
6.96 x 108
274.13
27.95
Mercury
3.18 x 1023
2.43 x 106
3.59
0.37
Venus
4.88 x 1024
6.06 x 106
8.87
0.90
Earth
5.98 x 1024
6.38 x 106
9.81
1.00
Moon
7.36 x 1022
1.74 x 106
1.62
0.17
Mars
6.42 x 1023
3.37 x 106
3.77
0.38
Jupiter
1.90 x 1027
6.99 x 107
25.95
2.65
Saturn
5.68 x 1026
5.85 x 107
11.08
1.13
Uranus
8.68 x 1025
2.33 x 107
10.67
1.09
Neptune
1.03 x 1026
2.21 x 107
14.07
1.43
Pluto
1.40 x 1022
1.50 x 106
0.42
0.04

5. Free Fall Acceleration
Since acceleration due to gravity is a constant 9.8 m/s2 downwards, what does this mean for velocity and distance?
Imagine I throw a ball upwards with a velocity of 20 m/s, what happens to the ball?
If I throw the ball up (positive direction) why does the ball eventually stop at the top and then proceed to drop?
Acceleration is acting on the ball.

6. Free Fall Acceleration
As the ball is moving up (positive), acceleration due to gravity (9.8 m/s2) is acting against it (negative).
Therefore, as the ball moves up, every second the velocity slows down by 9.8 m/s.
OHHHHH, so how long does it take to reach the top?

7. How long does it take to reach the top?
vf =vi +at
vf =0, vi = 20.0 m/s, a=-9.80 m/s2, t=?
0 =20.0 m/s +(-9.80 m/s 2)(t)
t = m/s m/s 2
t = seconds. 

8. Free Fall Acceleration
How high does that ball go?
d = vi t + ½ at2
vi =20.0 m/s, a=-9.80 m/s2, t=2.00 s, d = ?
d = (20.0 m/s)(2.04) + ½ (-9.80 m/s2)(2.04 s)2
d = (-20.39)
d = m 
This is only half of the motion of the ball, the ball then comes back down to my hand.
Thus the total distance travelled is 40.8 m, the displacement is 0 m, and the total time travelled is s.

9. Free Fall Acceleration v vs t graph
I threw the ball up at 20 m/s, everything left of 2.04 seconds indicates the ball being thrown upwards
Once the velocity is 0 m/s the ball has reached its max height
Everything right of 2.04 seconds indicates the ball moving downwards
Note the ball slows down in a positive direction (tossed up) as it approaches s, then it speeds up after the 2.04s in the negative direction (falls down)
The slope is constant at -9.8 m/s2, we know this because the v vs t line is straight.

10. Free Fall Acceleration d vs t graph
Ball starts in my hand (o m), is tossed up at 20 m/s and reaches m in 2.04 seconds.
At 2.04 seconds the ball reaches its maximum height.
Then the ball falls down back to my hand.
The slope is constant at -9.8 m/s2, we know this because the v vs t line is straight.

11. Free Fall Acceleration
Question, once the ball reaches 0 m/s does it fall down?
Acceleration due to gravity pulls it down at 9.8 m/s2
What would happen if, when the ball reaches 0 m/s and the acceleration was 0 m/s2?
The ball would not move, it would be in suspension and hovering.

12. Ex.
A student drops their homework down a wishing well. After 2.4 s it hits the water at the bottom. How deep is the well?
d = vi t + ½ at2
d = (0)(2.4s) + ½ (-9.8m/s2)(2.4s)2
d = 0 – m
d = m

13. Example vf = vi +at 0 = 15.0 m/s + (-9.80 m/s2)(t) t = -15.0 m/s
A football is kicked straight up in the air at 15.0 m/s.
How high does it go?
What is its total hang time?
Solution a)
b) t total = ( s)(2)
= seconds
= 3.06 seconds
vf = vi +at
0 = 15.0 m/s + (-9.80 m/s2)(t)
t = m/s
-9.80 m/s2
t = …. S
Or vf2 = vi2 – 2ad
d = vi t + ½ at2
d = (15.0 m/s)(1.5306… s) +½(-9.80m/s2)(1.5306…s)2
d = … m – m
d = m
d = 11.5 m

14. Homework
Pg 78, Q’s 41 – 45
Handout on uniform acceleration