free fall ch 2.5 wod are underline. a special case of constant acceleration. the most common case of...

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Free Fall Ch 2.5 WOD are underline. •A special case of constant acceleration. •The most common case of constant acceleration.

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Page 1: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free FallCh 2.5

WOD are underline.

•A special case of constant acceleration.

•The most common case of constant acceleration.

Page 2: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free Fall

• Really, just a special case of one dimensional kinematics.

• Uses the same equations.– But, since displacement is vertical, it is written as y

(instead of x).– Acceleration is no longer a variable, it is a fixed

value 'g'.

Page 3: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free FallFree FallFree Fall: The only acceleration or force is due : The only acceleration or force is due to a constant gravity.to a constant gravity.•All objects moving under the influence of “only gravity” are said to be in free fall. • Free fall acceleration does not depend on the object’s original motion, or on the mass.• Acceleration is the one thing constant to all falling objects. It is constant for the entire time they are in the air.

Page 4: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free FallFree FallFree Fall: The only acceleration or force is due : The only acceleration or force is due to a constant gravity.to a constant gravity.•This means that the significant figures of force due to other forces like Air Resistance are InSignificant.

Page 5: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Freely Falling Objects• A freely falling object is any object moving under the influence

of gravity alone.• Acceleration does not depend upon the initial motion of the

object!!!!!!!Examples of Freefall with different initial motions.– Dropped –released from rest– Thrown downward– Thrown upward– Throwing a ball Up or DOWN AFTER the ball leaves the hands.

Examples of NonFreefall or NOT FREEFALL.– Throwing a ball Up or DOWN BEFORE the ball leaves the

hands, while the hands are still pushing on it is NOT freefall.– A rocket engine is still burning and pushing up on a rocket.

Page 6: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Freely Falling Objects• A freely falling object is any object moving under the influence

of gravity alone.• Acceleration does not depend upon the initial motion of the

object!!!!!!!Examples of Freefall with different initial motions.– Dropped –released from rest– Thrown downward– Thrown upward– Throwing a ball Up or DOWN AFTER the ball leaves the hands.

So what does initial motion affect then???It changes the vi in vf = vi + at. It does not change the a.

Page 7: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Acceleration of Freely Falling Object

• The acceleration of an object in free fall is directed downward, regardless of the initial motion

• The magnitude of free fall acceleration is g= 9.80 m/s2 which is the number used in your textbooks, so use 9.80 so that you match the black and white textbooks.

• Only valid on the Earth’s surface at OUR ALTITUDE.• Not to be confused with g for grams

• British textbooks use 9.81 m/s2 instead of 9.80 m/s2 like American textbooks. Why?

Page 8: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Acceleration of Freely Falling Object

• The acceleration of an object in free fall is directed downward, regardless of the initial motion

• The magnitude of free fall acceleration is g= 9.80 m/s2 which is the number used in your textbooks, so use 9.80 so that you match the black and white textbooks.

• Only valid on the Earth’s surface at OUR ALTITUDE.• Not to be confused with g for grams

• GRAVITY VARIES, because of what?

Page 9: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Strength of gravity

• The value of 'g' is affected by altitude.

• At the top of Mt. Everest, g = -9.77 m/s2

(.028 % less)

Page 10: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Acceleration of Free Fall, cont.

• We will neglect air resistance• Free fall motion is constantly accelerated motion in

ONLY one dimension, which is y. DO NOT USE ax = -9.80 m/s2

• Let upward be positive• Use the kinematic equations with ay= -g= -9.80 m/s2

Page 11: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Acceleration of Free Fall, cont.

• Use the kinematic equations with a= g= -9.80 m/s2

REMEMBER: + means up, - means down• Write these equations on page: Kinematics Equation Page (Choose 1 side or other.) Y Axis y = yi + vi t - ½ gt2

vf = vi

- g t

vf2 = vi

2 -2g Δy

g= 9.80 m/s2

(Because a = g)

Y Axis y = yi + vi t + ½ at2

vf = vi

+ a t

vf2 = vi

2 + 2a Δy

g= -9.80 m/s2

(Because g = Neg)

Page 12: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free Fall –an object dropped

• If I drop a soccer ball off the roof of a very, very tall building, what is the velocity 8 seconds later?

• What is the displacement?

Solve these in your powerpoint notes.

Page 13: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free Fall –an object dropped

Vf = vi

- g t

V8 = 0 - (9.80 m/s2)(8 s)

V8 = -78.4 m/s

What is the displacement?y8 = yi + vi t - ½ gt2

y8 = 0 + 0 (8) - ½ (9.80 m/s2)(8 s)2

y8 = -313 m

What’s wrong with these answers?

vf = vi

+ a t

v8 = 0 + (-9.80 m/s2)(8 s)

v8 = -78.4 m/s

What is the displacement?y8 = yi + vi t + ½ at2

y8 = 0 + 0 (8)+½ (-9.80 m/s2)(8 s)2

y8 = -313 m

If I drop a soccer ball off the roof of a very, very tall building, what is the velocity 8 seconds later?

Page 14: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free Fall –an object dropped

Vf = vi

- g t

V8 = 0 - (9.80 m/s2)(8 s)

V8 = -78.4 m/s

What is the displacement?y8 = yi + vi t - ½ gt2

y8 = 0 + 0 (8) - ½ (9.80 m/s2)(8 s)2

y8 = -313 m

What’s wrong with these answers? Sig Fig. Ans: -80 and -300

vf = vi

+ a t

v8 = 0 + (-9.80 m/s2)(8 s)

v8 = -78.4 m/s

What is the displacement?y8 = yi + vi t + ½ at2

y8 = 0 + 0 (8)+½ (-9.80 m/s2)(8 s)2

y8 = -313 m

If I drop a soccer ball off the roof of a very, very tall building, what is the velocity 8 seconds later?

Page 15: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free Fall of an object thrown upward• Initial velocity is upward, so

positive• The instantaneous velocity

at the maximum height is zero.

• Vy=max = 0

• ay= -g = -9.80 m/s2 everywhere in the motion:

at the start, top, and end And everywhere inbetween.

vi≠0ay = -g

At Topvy = 0ay = -g

Page 16: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free Fall -object thrown upward

Vi > 0 ay = -g

At Topvy = 0ay = -g

A soccer ball is kicked directly upwards with a velocity of 8 m/s.

At what time will it’s velocity be zero?

At what time will its height peak?

What’s the highest it will go?

How much time will it spend in the air before it hits the ground?

With what velocity will it strike the ground?

Page 17: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free Fall -object thrown upward

Vi > 0 ay = -g

At Topvy = 0ay = -g

A soccer ball is kicked directly upwards with a velocity of 8 m/s.

At what time will it’s velocity be zero?Ans: vf

= vi + a t t=.816s = .8s

At what time will its height peak?Ans: Same Question. Notice vy = 0At the top.

What’s the highest it will go?Ans: yf = yi + vi t + ½ at2 3.27m = 3mOr use vf

2 = vi2 + 2a Δy

How much time will it spend in the air before it hits the ground? Double 1st anst = 1.6s

With what velocity will it strike the ground? Discuss ball, bullets, arrows, and air resistance.

Page 18: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

thrown upwards, free fall only

• The motion may be symmetrical IF it lands at the same height same height as thrown.

• Then tup= tdown

• Then vf = -vi

• Break the motion into two parts• up and down

Page 19: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free Fall –an object thrown downward

• ay= -g= -9.80 m/s2

• Initial velocity ≠0• Since up is positive, initial velocity will be

negative.Vo< 0a = -g

Page 20: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free Fall --object thrown DownwardSame Equations, but Vo will be

negative

A person fires a handgun straight downwards from a helicopter. The muzzle velocity of the gun is 300 m/s. The helicopter is 250 m above the ground.

With what velocity will the bullet hit the ground?

Page 21: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Free Fall --object thrown DownwardSame Equations, but Vo will be negative

A person fires a handgun straight downwards from a helicopter. The muzzle velocity of the gun is 300 m/s. The helicopter is 250 m above the ground.

With what velocity will the bullet hit the ground?

ANS: vf2 = vi

2 + 2a Δy = -308 m/sAll vi ,a, Δy = negative vf = choose neg sq.root.

Page 22: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Problem Solving tips

• Think about and understand the situation• GIVENS: List the given information. REMEMBER g = -9.80 m/s2

and at the top of the arc, vy = 0 (y direction only.)

• DIAGRAM: Make a quick drawing of the situation• FIND: List the unknown you are solving for.• EQUATIONS: List the equations used to solve.• CALCULATIONS: Show your work. Think about units• SOLUTION: List your answer.• Focus on the expected result• Think about what a reasonable answer should be

Page 23: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Example

We throw a bowling ball upwards from the edge of the roof.

What happens at points B, C and E?

Page 24: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Example

• Initial velocity at A is upward (+) and acceleration is -g(-9.80 m/s2)

• At B, the velocity is 0 and the acceleration is -g(-9.80 m/s2)

• At C, the velocity has the same magnitude as at A, but is in the opposite direction

• The displacement is –50.0 m (it ends up 50.0 m below its starting point)

Page 25: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Problem Solving –Question: If a flea can jump .440 m high, what is the initial velocity as it leaves the ground?

Page 26: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Problem Solving –Question: If a flea can jump .440 m, what is the initial velocity as it leaves the ground?

y = yi +vi t + ½ at2

v = vi + a t

vf2 = vi

2 +2ay

Page 27: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Problem Solving –Question: If a flea can jump .440 m, what is the initial velocity as it leaves the ground?

Just look at flea on the way up. y is .440 m vi is what we’re looking for

a is -9.8 m/s2 vf is zero, since we’re looking at the top of his jump.

Page 28: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Problem Solving –Question: If a flea can jump .440 m, what is the initial velocity as it leaves the ground?

Just look at flea on the way up. y is .440 m vi is what we’re looking for

a is -9.8 m/s2 vf is zero, since we’re looking at the top of his jump.

vf2 = vi

2 +2ay

Page 29: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Problem Solving –Question: If a flea can jump .440 m, what is the initial velocity as it leaves the ground?

0 = vi 2 + 2 (-9.80m/s2)(.440m)

vf2 = vi

2 +2ay

Page 30: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Problem Solving –Question: If a flea can jump .440 m, what is the initial velocity as it leaves the ground?

vf2 = vi

2 + 2 a y0 = vi

2 + 2 (-9.80m/s2)(.440m) = vi

2 - 8.62 m2/s2

so vi = + / - 2.94 m/s

Which makes sense, + or -?

Page 31: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Problem Solving –Question: If a flea can jump .440 m, how much time does he spend going up?

yf = yi +vi t + ½ at2

vf = vi

+ a t

vf2 = vi

2 +2ay

Page 32: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Problem Solving –Question: If a flea can jump .440 m, how much time does he spend going up?

a is -9.8 m/s2 vi = 2.94 m/s vf = 0

-9.8 m/s2 = (0 - 2.94 m/s) / t t = .300 sec

vf = vi

+ a t

Page 33: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Problem Solving –What is the velocity when he lands?

Just look at the problem on the way down.

vf2 = 0 + (-9.80m/s2)(-.440m)

= 0 + 8.62 m2/s2

vf = +/- 2.94 m/s

Which makes sense, + or - ?

vf2 = vi

2 +ay

Page 34: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Problem Solving

• How much time does he spend in the air?

yf = yi +vi t + ½ at2

vf = vi

+ a t

vf2 = vi

2 +2ay

Page 35: Free Fall Ch 2.5 WOD are underline. A special case of constant acceleration. The most common case of constant acceleration

Problem Solving

• How much time does he spend in the air?

• -2.94m/s = 2.94m/s + (-9.81m/s2)*t• -5.82m/s = -9.81m/s2*t • t = .600 sec

• Twice the time it took to go up

vf = vi

+ a t