Projectile Motion Lecture Notes: PHYS 100, Lecture 4, Study notes of Physics

Download Projectile Motion Lecture Notes: PHYS 100, Lecture 4 and more Study notes Physics in PDF only on Docsity!

PHYS 100: Lecture 4

PROJECTILE MOTION

1.4 1.2^1 0.8 0.6 0.4 0.2^00

(^10 15) x(m)

y(m)^ y = (v/v) x – (g/2v^0 T

22 )xT

v^0 Velocity of TrainvT

Music

Who is the Artist? Who is the Artist?A)^ Sarah Vaughn A)Sarah VaughnB)^ Billie Holiday B)Billie HolidayC)^ Dinah Washington C)Dinah WashingtonD)^ Nina Simone D)Nina SimoneE)^ Ella Fitzgerald E)Ella Fitzgerald

Why?Why?

Last time you picked her when I played Chet Baker. I Last time you picked her when I played Chet Baker. Ifigured you should hear her!^ figured you should hear her!A true classic..^ “dramatic, defiant, and A true classic..“dramatic, defiant, and

uncategorizable”uncategorizable”

Important figure in the 60^ Important figure in the 60

’s…’s…

If you havenIf you haven’’t heard her before, you really shouldt heard her before, you really should

THE BIG IDEASTHE BIG IDEAS

NOTE: THE BIG IDEAS ARE ALWAYS GIVEN IN THE LAST SLIDE 1.^ Motion described as 1-D freefall in one frame is described as“projectile motion” in all other frames.2.^ Description of projectile motion is superposition of x and y motionsa)^ Constant velocity in x (horizontal)b)^ Constant acceleration in y (vertical)

Ball on a CartBall on a Cart DEMO:^ Ball is ejected straight up from a cup in a cart moving at constant velocity.When it falls back , will it land

(A) in back of the cup (B) in the cup (C) in front of the cup B^ v A^ C This is^ THE POINT:^

Descriptions of same motion from two different reference frames^ gtvv −=^ y^00 = vx

v^021 gttvy^ −=^20 0 = x

Cart frame:^ motion is 1-D free fall: ball goes up, stops, and comes straight back down

yC^ xC

gtvv −= y 0 vv = x

21 gttvy −= 20 vtx =

Ground frame:^ motion is 2-D projectile: ball follows parabolic trajectory

xG yG

Preflight 1Preflight 1 Without air resistance, an object dropped from a plane flying at constant speedin a straight line will: (A) Quickly lag behind the plane(B) Remain vertically under the plane You said: • It won't remain under the plane or go ahead of itbecause the plane is flying away from the falling object,while the object is falling straight down. •The plane and the object have the same horizontalvelocity. Much like coffee stays in the same positionrelative to the plane once it is poured, the same willhappen for the object. • The object is being dropped with a velocity that hascomponents in the x and y directions. The y directionhas a velocity starting at zero but increasing due to thepull of gravity. The x direction is the velocity at whichthe plane is traveling when the object is dropped.

Move ahead of the plane(C) (^) (D) It depends on the object^706050403020100 A^ B Same as Ball on Cart !!^ C^ D

Practice (with a Point)Practice (with a Point) • How do we calculate how long the ball stays in the air? • At the top (1/2^ of the time in the air)

the velocity of the ball is zero

• The velocity is being reduced by

gtv^ g^ m/sec every second Two objects are thrown straight up from the same height. Object 1 is given aninitial velocity vwhile Object 2 is given an initial velocity 2v^0 v^ −=^01

Which object is in the air the longest?^ (A) Object 1^

(B) Object 2^ (C) Same amount of time^ 2v^0 v^01

gtvv −= (^202) v^0 t = (^) top^ g

v (^20) t = topg v 2v (^0) v 0 t^1

More PracticeMore Practice

-^ How do we calculate how long the ball stays in the air?^ • At the top^ (1/

of the time in the air)

the y-component of the velocity of the ball is zero• The y-component of the velocity is being reduced by

g^ m/sec every second Two objects are thrown with different initial velocities from the same height. Object 1is thrown with v

o^ at angle θ = 60while Object 2 is thrown with 1.5v 0

oat angle θ = 30 0

Which object is in the air the longest?^ (A) Object 1^

(B) Object 2^ (C) Same amount of time^ v1.5v^0

(^0) oo (^30601 ) v^ °^60 sin^0 t = (^) top^ g^ gtvv −°= 60 sin)( y^01

gtv v^ °^30 sin (^5). (^1 0) t = (^) top^ g v −°= 30 sin (^5). (^1) )( y (^02) o^ Sin60=^ √3/2 = .87^2 √3 o^601

o^ Sin30=^ 1/2 = .50o^ 1.5 Sin30=^ 3/4 =. vy o 1.5vsin30 0 t o vsin60 0 12

The PointThe Point What determines how long an object will stay in the air when thrown?^ (A)^ Only the magnitude of the initial velocityOnly the horizontal component of the initial velocity(B)(C)Only the vertical component of the initial velocityBoth the horizontal & vertical component of the initial velocity(D) • How do we calculate how long the ball stays in the air? • At the top^ (1/2^ of the time in the air)

the y-component of the velocity of the ball is zero• The y-component of the velocity is being reduced by

g^ m/sec every second

gtvv −= θsin y 0

v^ θsin^0 t = topg

How High?How High?

-^ How do we calculate how high the ball goes?^ • At the top^ (1/

of the time in the air)

the y-component of the velocity of Two objects are thrown with different initial velocities from the same height. Object 1is thrown with vat angle^0 the ball is zero• Find the time when this occurs (we’ve done that a lot today)• Determine the height at this time.

o^ θ = 60while Object 2 is thrown with 1.5v

oat angle θ = 30 0

Which object goes the highest?^ (A) Object 1^

(B) Object 2^ (C) Same amount of time^ v1.5v^0

(^0) oo (^30601 ) v^ °^60 sin^0 t = (^) top^ g^ gtvv −°= 60 sin)( y^01

gtv v^ °^30 sin (^5). (^1 0) t = (^) top^ g v −°= 30 sin (^5). (^1) )( y (^02) THE MATH:^21 sin^ gttvy −=^ θ^0 2 v^ θsin^0 t^ =^ g

(^20012) (^0) max

 sinsinsin −

=^

vvggg

vy

22 v^ θ sin 01 y −=max 2 g

Preflight 5Preflight 5 A battleship simultaneously fires twoshells at enemy ships.^ If the shellsfollow the parabolic trajectoriesshown, which ship gets hit first? (A) AB(C)^ (D) more info (B) At the same time You said: • The vertical distance that shell A travels looks likemore than 2x of B's, but the distance B's travels looksless than 2x so they would take the same time. • Ship B would get hit first because the top hight ofthe A shell is higher than the B shell. This verticleposition is the only thing that determines how long itwill take for the shell to come down. • We need to know the initial velocities of both shotsand the distances of ship A and ship B.

50 40 30 20 10 0 A^ B^ C^

D

SKIP to SOLUTION

Preflight 5Preflight 5 A battleship simultaneously fires twoshells at enemy ships.^ If the shellsfollow the parabolic trajectoriesshown, which ship gets hit first? (A) AB(C)^ (D) more info (B) At the same time WHAT HAVE WE LEARNED ABOUT PROJECTILE MOTION TODAY ?? v^ θsinThe time in the air is determined by the (^0) t = (^) top vertical component of the initial velocity^ g^22 The maximum height is determined by the v^ θ^ sin^01 y −= (^2) maxvertical component of the initial velocity^ g^ A goes^ HIGHER

so A takes longer !!