One dimensional motion

2.3 free fall

Objectives

1. Relate the motion of a freely falling body to motion with constant acceleration.

2. Calculate displacement, velocity, and time at various points in the motion of a freely falling object.

3. Compare the motion of different objects in free fall.

Homework due: 10/4Test: 10/4

Key words and ideas

Notes

Free fall

Text Page 60 Figure 2.14• If the feather and apple experiment were

performed on the moon, where free-fall acceleration is approximately on-sixth the value of free-fall acceleration on Earth, how would the picture compare with figure 2-14?

• If the initial velocity of the ball is 10.5 m/s when it is thrown from the ground, what is the final velocity of the ball at the end of its flight?

Text Page 61 Figure 2.15

• How do you describe the motion from this graph?

• At what time is the velocity zero?

• At what time is the acceleration zero?

Text Page 61 Figure 2.16

Two important motion characteristics that are true of free-falling objects

• Free-falling objects do not encounter air resistance.

• All free-falling objects (on Earth) accelerate downwards at a constant rate.

“g” - The “Magic” Number

• “g” means ACCELERATIONACCELERATION DUE TO GRAVITY

• Each planet, star, moon, or other large object has its own value for “g”

Examples • “g” is 1.62 m/s2 on the Moon• “g” is 26 m/s2 on Jupiter• “g” is 9.81 m/s2 on Earth

If the feather and elephant experiment were performed on the moon, would they still fall at the same rate, just like on Earth?

• Dropped from rest

• Throwing downward

• Throwing upward

• Throwing side ways

a = -g = -9.81m/s2 for all free fall objects.

(the ‘-’ sign means “downward”)

• Since free fall motion has constant acceleration, we can apply all kinematics equations on free fall motion.

Free fall – motion with constant acceleration a = -g

To solve free fall motion problems, we can use kinematics equations with constant acceleration.

a = -g∆x = ∆y

Velocity of a dropped object• An object falls from rest. What is its velocity at the end

of one second? Two seconds? Three seconds?

t = 0 vi = 0 m/s

t = 1 s v = -9.81 m/s

t = 2 s v = -19.62 m/s

t = 3 s v = -29.43 m/s

atvv if a = -g = -9.81 m/s2 (‘-’ means “downward”)

Displacement of a dropped object• An object falls from rest. How far has it

fallen at the end of one second? Two seconds? Three seconds?

t = 0 d = 0 m

t = 1 s d = -4.905 m

t = 2 s d = -19.62 m

t = 3 s d = -44.145 m

2

2

1attvd i

+

Velocity and distance of a free falling object dropped from rest

Throwing Downward• An object is thrown downward from

the top of a 175 meter building with an initial speed of 10 m/s.

– What acceleration does it experience?

• -9.81 m/s2

– What is the object’s initial velocity?• -10 m/s

vi = -10 m/sa = -9.81 m/s2

“a” and “vi” in SAME DIRECTION

Building Example (cont.)

• What is the object’s velocity as it hits the ground?

• vi = -10 m/s

• d = -175 m• a = -9.81 m/s2

• vf = ?

• How long does it take the object to hit the ground?

• t = ?

vf2 = vi

2 + 2advf

2 = (10m/s)2 + 2(-9.81 m/s2)(-175 m)vf = -59.4 m/s

vf = vi + at-59.4 m/s = -10 m/s + (-9.81 m/s2) t

t = 5.05 s

Throwing Upward

• A cannon fires a shot directly upward with an initial velocity of 50 m/s.

– What acceleration does the cannonball experience?

• -9.81 m/s2

– What is the cannonball’s initial velocity?

• +50 m/s

t = 0 svi = +50 m/s

t = 1 sv = +40.19 m/s

t = 2 sv = +30.38 m/s

“a” and “vi” in OPPOSITE DIRECTION

atvv if

Cannon (cont.)• What is the object’s velocity as it reaches the

top of its flight?

• How long does it take the cannonball to reach the top of its flight?

• What is the maximum height of the cannonball?

0 m/s (All objects momentarily STOP at the top of their flight)

vf = vi + at0 = +50 m/s + (-9.81 m/s2) t

t = 5.1 s

d = vit + ½ at2

d = (+50 m/s)(5.1 s) + ½ (-9.81 m/s2)(5.1 s)2

d = +127.4 m

• A rocket is fired straight up into the air to an altitude of 20 m.

1.At the top of its flight, what is its velocity? .

2.At the top of its flight, what is its acceleration?

3.After the rocket falls back to its launching pad, what is its displacement?

At the top of an object’s flight, its velocity is zero. Why its acceleration is not zero?

Page 63 - Sample problem 2F• Jason hits a volleyball so that it moves with an

initial velocity of 6.0 m/s straight upward. If the volleyball starts from 20 m above the floor, how long will it be in the air before it strikes the floor? Assume that Jason is the last player to touch the ball before it hits the floor.

Class work

• Page 64 – practice 2F

Free Fall Graphs of a dropped object

A position versus time graph for a free-falling object

A velocity versus time graph for a free-falling object

velocity increase in the negative direction – slope is neg. & increasing

velocity increases in the negative direction at constant rate – slope is neg. & constant

t

Dis

p.

t

velo

city

time

velo

city Upward: velocity is big, positive, decreasing,

slope is constant (a = -9.8 m/s/s).

Top, velocity is zero. Slope remains the same (acceleration is still -9.8 m/s/s)

Downward: velocity increases in negative direction at the same constant rate of -9.81 m/s/s (slope remains the same), reaches the same speed as it started upward.

Free Fall Graphs of upward object

time

pos

itio

n Upward: displacement increases, slope is positive, decreasing (velocity is positive, decreasing)

Top: slope = 0 (its velocity is zero)

Downward: displacement decreases, its slope increases in negative direction (velocity is negative, increasing)

Review1. An object accelerates uniformly from rest to a speed of

50. m/s in 5.0 seconds.  What is the average speed of the object during the 5.0 second interval?

2. A car travels 90. meters due north in 15 seconds. Then the car turns around and travels 40. meters due south in 5.0 seconds. What is the magnitude of the average velocity of the car during this 20.-second interval?

3. A locomotive starts at rest and accelerates at 0.12 meters per second squared to a speed of 2.4 meters per second in 20. seconds. Describe its velocity and acceleration using words “constant” and “increasing”.

4. A race car traveling at 10. m/s accelerates at a rate of 1.5 m/s2 while traveling a distance of 600. meters.  What is the final speed of the race car?

do now

• What is free fall? (2 conditions)

Which pair of graphs represents the same motion?

A.

C.

B.

D.

• A cart travels with a constant nonzero acceleration along a straight line. Which graph best represents the relationship between the distance the cart travels and time of travel?

A.

B.

C.

D.

• A car traveling at a speed of 13 meters per second accelerates uniformly to a speed of 25 meters per second in 5.0 seconds.

• A truck traveling at a constant speed covers the same total distance as the car in the same 5.0- second time interval. Determine the speed of the truck.

• Answer: ______________ m/s

• A skier starting from rest skis straight down a slope 50. meters long in 5.0 seconds. What is the magnitude of the acceleration of the skier?

A. 20. m/s2

B. 9.8m/s2

C. 5.0 m/s2

D. 4.0 m/s2

• When velocity is zero and acceleration is negative, what happens to the object’s motion?

1. The object slows down.

2. The object speeds up from a negative velocity.

3. Nothing happens to the object.

4. The object speeds up from rest.

10/2 do now

• A trucker drives with an average velocity of 27 m/s toward the east on a perfectly straight highway. What will the trucker’s displacement be after 2.0 h?

1. 97 km to the east

2. 1.6 × 102 km to the east

3. 1.9 × 102 km to the east

4. 3.2 × 102 km to the east

• Frame of reference– Toward each other – add speed– Move in the same direction – subtract speed

• Vector, scalar– Vector has magnitude (number) and direction– Scalar has magnitude only

• Distance, displacement (x, y, d, ∆x, ∆y)– Distance – scalar, length of the path– Displacement – vector, direct distance

• Speed, velocity, average velocity, instantaneous velocity (v, v)– vavg = ∆x/∆t; vavg = d/t; Unit: m/s– v = slope of d vs. t graph;

• Acceleration (a)– aavg = ∆v/∆t = (vf-vi)/∆t, unit: m/s2

– aavg = slope of v vs. t graph;– Disp. = area under v vs. t graph;

• Free fall - two conditions: (g = 9.81 m/s2)– No air resistance, all objects falls the same distance during same time– a = - g; – tup = tdown; vup = -vdown; – at top: v = 0, a = -g

d

t

d

t

v

t

v

t

v

td

t

v = d/t; v = ½ (vi + vf)

aavg = ∆v/∆t = (vf - vi)/∆t

vf = vi + a∆t

d = vit + ½ at2

vf2 = vi

2 + 2ad

Lab 5 – time interval of free fall• Objective: determine your reaction time• Material: ruler• Data table – show work for one trial:

trial ∆y (m)

1

2

3

4

5

Average

• conclusion: use your data to determine your reaction time: use equation, substitute number and units and answer with units.

Do this lab with the person sits next to you

Do now• A trucker drives with an average

velocity of 27 m/s toward the east on a perfectly straight highway. What will the trucker’s displacement be after 2.0 h?

a. 97 km to the east b. 1.9 102 km to the east c. 1.6 102 km to the eastd. 3.2 102 km to the east

Lab 3 – Position vs. Time graph (5 pt.)• Purpose: determine the position-time graph of an electric

car moving at constant velocity (5 pt)• Material: electric car, ticker tape, spark timer, ruler (5 pt)• Procedure: briefly describe how you measured your data so

that others will know to follow your steps to complete this lab (10 pt)

• Data table: (20 pt) Time (s) Distance

(cm)

0 0

0.1

0.2

0.3

0.4

0.5

• Data Analysis: graph data; determine slope – must show work; what is the meaning of your slope? (30 pt)

Once you finish your lab, you may correct lab 2 and do problems in your guided notes – page 11 & 12