Free fall

Free Fall

Free fall – motion under the influence of the gravitational force only (neglects air resistance)

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Elapsed time

time that has passed from the beginning of a fall

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Gravity and Free fall

Acceleration due to gravity is 9.8 m/s2, downward

Every second that an object falls, the velocity increases by 9.8 m/s

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Sample 1If an object is dropped from rest at the top of a cliff, how fast will it be going after 1 second?

2 seconds? 10 seconds?

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First, Identify the GIVENS?When an object is falling, assume that

a = 9.8 m/s2

“from rest” tells us that Vi = 0

“how fast will it be going?” is asking us to find Vf = ? (this is the unknown)

the first problem gives us a time of t = 1 sWork this on YOUR paper using GUESS…

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Therefore…G: a = 9.8 m/s2

Vi = 0

t = 1 s

U: Vf = ?

E: Vf = Vi + at

S: Vf = 0 + (9.8 m/s2) (1 s)

S: Vf = 9.8 m/s

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Now find out how fast it will be falling after 2 s…G: a = 9.8 m/s2

Vi = 0

t = 2 s U: Vf = ?

E: Vf = Vi + at

S: Vf = 0 + (9.8 m/s2) (2 s)

S: Vf = 19.6 m/s

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Now find out how fast it will be falling after 10 s…G: a = 9.8 m/s2

Vi = 0

t = 10 s U: Vf = ?

E: Vf = Vi + at

S: Vf = 0 + (9.8 m/s2) (10 s)

S: Vf = 98 m/s

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Calculating How Fast Free Fall Is

If we use Vf = Vi + at, and Vi = 0 then we actually have an equation that reads… Vf = at, where a = 9.8 m/s2. When a = 9.8 m/s2, we call that constant g. Therefore, we can use the following new, or derived equation… Instantaneous speed = acceleration X elapsed time

v = gt 9

Sample Problem #2

The Demon Drop ride at Cedar Point Amusement Park falls freely for 1.5 s after starting from rest.What is its velocity at the end of 1.5 s?How far does it fall?

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What is its velocity at the end of 1.5 s?

G: a = 9.8 m/s2

Vi = 0 (starting from rest)

t = 1.5 sU: Vf = ?

E: Vf = Vi + at

S: Vf = 0 + (9.8 m/s2) (1.5 s)

S: Vf = 14.7 m/s

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How far does it fall?G: a = 9.8 m/s2

Vi = 0

t = 1.5 s Vf = 14.7 m/s (from previous part)

U: d = ? (how far)E: d = Vit + ½ at2

(use any of the 3 motion formulas

with d in them)

S: d = (0)(1.5s) + ½ (9.8 m/s2 )(1.5s)2

S: d = 11.03 m 12

Calculating How Far Free Fall Is

If we use d = Vit + ½ at2 , and Vi = 0 then we actually have an equation that reads… d = ½ at2 , where a = 9.8 m/s2. When a = 9.8 m/s2, we call that constant g. Therefore, we can use the following new, or derived equation…

d = ½ gt2

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Acceleration due to Gravity

If the object is moving down, a = 9.8 m/s2

If the object is moving up, a = - 9.8 m/s2

a = 9.8 m/s2 (speeding up)

a = - 9.8 m/s2 (slowing down)

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Throwing an object into the air

When an object is thrown into the air, the velocity at its highest point is ZERO!!!

Begin

Vi = ?

Begin

Vi = 0 m/s

End

Vf = 0 m/s

End

Vf = ?

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Sample Problem #3A ball is thrown vertically into the air with an initial velocity of 4 m/s.

How high does the ball rise?How long does it take to reach its highest point?

If the ball is caught in the same spot from which it was thrown, what is the total amount of time that it was in the air?What is its velocity just before it is caught?

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How high does the ball rise?

HINT: draw a picture and label it

Begin

Vi = 4 m/s

End

Vf = 0 m/s

d= ?

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How high does the ball rise?

G: a = -9.8 m/s2 (notice the negative sign, ball moving upward)

Vi = 4 m/s

Vf = 0 m/s (at the top, before it starts to fall, it stops)

U: d = ?E: 2ad = Vf

2 – Vi2, (solve for d) d = Vf

2 – Vi2

2aS: d = 02 – (4 m/s)2

2(-9.8 m/s2 )S: d = 0.816 m

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How long does it take to reach its highest point?

G: a = -9.8 m/s2 U: t = ? Vi = 4 m/s

Vf = 0 m/s

d = 0.816 mE: Vf = Vi + at, (solve for t) t = Vf – Vi

a

S: t = 0 m/s – 4 m/s

(-9.8 m/s2)

S: t = 0.408 s 19

If the ball is caught in the same spot from which it was thrown,

what is the total amount of time that it was in the air?

G: a = 9.8 m/s2 (ball going down, positive) U: t = ? Vi = 4 m/s

Vf = 0 m/s

d = 0.816 mE: d = Vit + ½ at2 ; derived to d = ½ gt2, solve for t…

t2 = (2d)/gS: t2 = (2)(0.816 m)/(9.8 m/s2) *don’t forget to take the square root

S: t = 0.408sTotal time = time Going up + time going downTotal time = 0.408s +0.408s = 0.816s 20

G: a = 9.8 m/s2 (ball moving down before it is caught)

Vi = 0 m/s

d = 0.816 m t = 0.408 sU: Vf = ?

E: 2ad = Vf2 – Vi

2, (solve for Vf) Vf2 = 2ad +

Vi2

S: Vf2 = 2(9.8 m/s2 )(0.816 m) + 02

S: Vf = 3.99 or 4 m/s

What is its velocity just before it is caught?

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You’re done!

Now try some problems on your own.

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