Acceleration



Free Fall

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Free Fall & Acceleration of Gravity

Gravity ( g ) – a mutual attraction between the mass of the Earth and the mass of an object in the

vicinity of the earth.

Free fall – the ideal falling motion of an object that is acted upon solely by the force of gravity.

Construct a statement regarding the motion of falling objects.

As objects fall, their speed increases (accelerate).

Assumed Characteristics of Objects in Free Fall

1. All objects accelerate downward at a rate of 9.81 m/s2

2. Free falling objects do not encounter air resistance.

If these factors differ, how would the rate of free fall be effected?

If the acceleration due to gravity (g) decreased, than the object would fall more slowly (acc. less).

If air resistance increased, the object would fall more slowly.

|Time (s) |Speed (m/s) |Distance (m) |

|0.00 |0 |0 |

|1.00 |9.81 | 4.91 |

|2.00 | 19.6 | 19.6 |

|3.00 | 29.4 | 44.1 |

|4.00 | 39.2 | 78.5 |

|5.00 | 49.1 | 123 |

Formula →

Sketch the distance fallen, speed and acceleration graphs for the falling ball. How would these change if displacement and velocity were graphed instead?

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| |Air Resistance |Gravity |

|Earth |----- |9.81 m/s2 |

|Moon |0 | 1.6 m/s2 |

|Deep Space |0 |0 |

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How does mass effect free fall?

Aristotle believed that the rate of fall for a body depended on its weight.

Galileo believed that in the absence of air, all bodies fall at the same rate.

|Demo #1 |Demo #2 |

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|Sketch: |Sketch: |

|[pic] | |

|Observations: | |

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|Book falls faster. | |

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| |Observations: |

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| |Both fall at same rate. |

|Video Clip #1 |Video Clip #2 |Video Clip #3 |

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|Sketch: |Sketch: |Sketch: |

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|Observations: |Observations: |Observations: |

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|Penny hits first. |Both hit as same time. |Both hit as same time. |

Gravity Problems

1. A ball is dropped down a shaft and hits the bottom in 3.2 seconds. Determine:

a) the depth of the shaft

d = vit + ½ at2 = d = ½ gt2

= ½ (9.81 m/s2)(3.2 s)2

= 51 m

b) how fast the ball is going when it hits the bottom (impact velocity)

vf2 = vi2 + 2ad

= 2(9.81 m/s2)(51 m)

= 31 m/s

2. A stunt man jumps off the Brooklyn Bridge which is 40. meters high. Determine:

a) the time it takes to hit the water

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b) his impact velocity

vf = vi + at

= (9.81 m/s2)(2.9 s)

= 31 m/s

c) How would these values change if he were on the moon?

vf = vi + at

= (1.6 m/s2)(7.1 s)

= 11 m/s

Throwing Up

A ball is thrown up into the air, as shown in the time-elapsed diagram. Each frame represents the position of the ball after 1.00 additional second of flight.

a) How long is it in the air?

6 seconds

b) How long did it take to get to the top of its path?

3 seconds

c) How fast was it going when it left the ground?

vf = vi + at = - (- 9.81 m/s2)(3.00 s) = 29.4 m/s

d) Draw and label an arrow at each location to represent the instantaneous velocity of the ball.

e) Draw and label an arrow at each location to represent the acceleration of the ball.

f) Fill in the chart below.

| |Height (m) |Velocity (m/s) |Acceleration (m/s2) |

|A |d = vit + ½ at2 |vf = vi + at |- 9.81 |

| | | | |

| |0 m |29.4 m/s | |

|B |d=(29.4m/s)(1s)+½ (-9.81m/s2)(1s)2 |vf = 29.4 m/s + (-9.81m/s2)(1s) |- 9.81 |

| | | | |

| |24.5 m |19.6 m/s | |

|C |d=(29.4m/s)(2s)+½ (-9.81m/s2)(2s)2 |vf = 29.4 m/s + (-9.81m/s2)(2s) |- 9.81 |

| | | | |

| |39.2 m |9.80 m/s | |

|D |d=(29.4m/s)(3s)+½ (-9.81m/s2)(3s)2 |vf = 29.4 m/s + (-9.81m/s2)(3s) |- 9.81 |

| | | | |

| |44.1 m |-.03 m/s = 0 m/s | |

|E |d=½ (9.81m/s2)(1s)2 |vf = (9.81m/s2)(1s) |9.81 |

| | | | |

| |44.1 m – 4.90 m = 39.2 m |9.81 m/s | |

|F |d=½ (9.81m/s2)(2s)2 |vf = (9.81m/s2)(2s) | 9.81 |

| | | | |

| |44.1 m – 19.6 m = 24.5 m |19.6 m/s | |

|G |d=½ (9.81m/s2)(3s)2 |vf = (9.81m/s2)(3s) | 9.81 |

| | | | |

| |44.1 m – 44.1 m = 0 m |29.4 m/s | |

From the example on the previous page, describe how each of the following changed throughout the flight of the ball:

|Speed |Velocity |Acceleration |Height |Displacement |Distance | |Going Up |Decreased |Decreased |Same |Increased |Increased |Increased | |Coming Down |Increased |Increased |Same |Decreased |Decreased |Increased | |

? ? ? ? ? ? ? ? ? ? ? ? ? ? Questions ? ? ? ? ? ? ? ? ? ? ? ? ? ?

1

1. A boy throws a ball straight up in the air with an initial velocity of 35.0 m/s. Determine:

a) the time it takes to get to the top of its flight

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b) highest point reached

d = vit + ½ at2 = (35.0 m/s)(3.57 s) + ½ (- 9.81 m/s2)(3.57 s)2

= 133 m – 64 m = 69 m

c) impact velocity

35.0 m/s

2. A football is punted straight up and remains airborne for 4.6 seconds. Determine:

a) the time it takes to get to the top of its flight

d) vertical launching velocity

vf = vi + at

= (- 9.81 m/s2)(2.3 s)

= 23 m/s

e) highest point reached

d = vit + ½ at2 = (23 m/s)(2.3 s) + ½ (- 9.81 m/s2)(2.3 s)2

= 53 m – 26 m = 27 m

-----------------------

g = - 9.81 m/s2

Equator = 9.75 m/s2

North Pole = 9.83 m/s2

Canada = 9.81 m/s2

What factors would effect g ?

d = ½ gt2

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"v = gt

acceleration

displacement

distance

velocity

speed

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Air

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Vacuum

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Moon

Givenion.DSMT4 [pic]

∆v = gt

acceleration

displacement

distance

velocity

speed

[pic]

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Air

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Vacuum

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Moon

Givens

t = 3.2 s

a = 9.81 m/s2

Unknowns

d = ?

Givens

d = 51 m

vi = 0 m/s

a = 9.81 m/s2

Unknowns

vf = ?

Givens

d = 40 m

vi = 0 m/s

a = 9.81 m/s2

Unknowns

t = ?

Givens

t = 2.9 s

vi = 0 m/s

a = 9.81 m/s2

Unknowns

vf = ?

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Do Now!

Givens

vi = 35 m/s

vf = 0 m/s

a = 9.81 m/s2

Unknowns

t = ?

Givens

t = 3.6 s

Unknowns

d = ?

Givens

vf = 0 m/s

t = 2.3 s

a = 9.81 m/s2

Unknowns

vi = ?

4.6 s / 2 = 2.3 s

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