University Physics: Mechanics

University Physics: Mechanics

Ch4. TWO- AND THREE-DIMENSIONAL MOTION

Lecture 5

Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com

5/2Erwin Sitompul University Physics: Mechanics

Homework 4: The PlaneA plane flies 483 km west from city A to city B in 45 min and then 966 km south from city B to city C in 1.5 h.From the total trip of the plane, determine:(a) the magnitude of its displacement;(b) the direction of its displacement;(c) the magnitude of its average velocity;(d) the direction of its average velocity;(e) its average speed.


Solution of Homework 4: The Plane

A

483 km, 45 min

B

C

966

km,

1.5

hΔr1

→ 1ˆ483i kmr

2ˆ966j kmr

1 45 min 0.75 ht

2 1.5 ht

total 1 2r r r ˆ ˆ483i 966j km

(a) the magnitude of its displacement2 2

total ( 483) ( 966)r

1080.021 km

(b) the direction of its displacement1

total966tan483

r

63.435 243.435 • Quadrant III

• Quadrant I

total 1 2t t t 0.75 1.5 2.25 h

Δr2→

243.435

AB

C

Δrto

tal→


(c) the magnitude of its average velocity

Solution of Homework 4: The Plane

(d) the direction of its average velocity

(e) its average speed

totalavg

total

rvt

ˆ ˆ483i 966j km

2.25 h

ˆ ˆ214.667i 429.333j km h

2 2avg ( 214.667) ( 429.333)v 480 km h

1avg

429.333tan214.667

v 243.435 • Quadrant III

avgs total distance traveled

total time483 km 966 km

2.25 h

644 km h


When a particle’s velocity changes from v1 to v2 in a time interval Δt, its average acceleration aavg during Δt is:

change in velocityaverage

acceleration time interval

2 1avg

v vat

v

t

If we shrink Δt to zero, then aavg approaches the instantaneous acceleration a ; that is:

dvadt

ˆ ˆ ˆ( i j k)x y zd v v vdt

ˆ ˆ î j kyx zdvdv dv

dt dt dt

Average and Instantaneous Acceleration→ →

→

→→


We can rewrite the last equation asˆ ˆ î j kx y za a a a

where the scalar components of a are:

,xxdv

adt

,yy

dva

dt z

zdvadt

Acceleration of a particle does not have to point along the path of the particle

Average and Instantaneous Acceleration

→


A particle with velocity v0 = –2i + 4j m/s at t = 0 undergoes a constant acceleration a of magnitude a = 3 m/s2 at an angle 130° from the positive direction of the x axis. What is the particle’s velocity v at t = 5 s? Solution:

0x x xv v a t 0y y yv v a t 0 2 m sxv 0 4 m syv

3 cos130xa 3 sin130ya 21.928 m s 22.298 m s

2 ( 1.928)(5)xv 11.64 m s

4 (2.298)(5)yv 15.49 m s

Thus, the particle’s velocity at t = 5 s is

ˆ ˆ11.64i 15.49j m s.v

Average and Instantaneous Acceleration^→

→

^→

At t = 5 s,


Projectile Motion Projectile motion: a motion in a vertical plane, where the

acceleration is always the free-fall acceleration g, which is downward.

Many sports involve the projectile motion of a ball. Besides sports, many acts also involve the projectile motion.

→


Projectile Motion Projectile motion consists of horizontal motion and vertical

motion, which are independent to each other. The horizontal motion has no acceleration (it has a constant

velocity). The vertical motion is a free fall motion with constant

acceleration due to gravitational force.

0xa

ya g

29.81m sg


Projectile Motion

0xa

ya g0 0 0

ˆ î jx yv v v

0 0 0cosxv v

0 0 0sinyv v


Projectile MotionTwo Golf Balls

• The vertical motions are quasi-identical.

• The horizontal motions are different.

0 0xv

0 0xv


Projectile Motion AnalyzedThe Horizontal Motion

0 0xx x v t

0 0 0( cos )x x v t

The Vertical Motion1 2

0 0 2yy y v t gt

1 20 0 0 2( sin )y y v t gt

0 0sinyv v gt 2 2

0 0 0( sin ) 2 ( )yv v g y y


The Horizontal Range0x x R

1 20 0 20 ( sin )v t gt

0 0( cos )R v t

0 0y y

Eliminating t,20

0 02 sin cosv

Rg

• This equation is valid if the landing height is identical with the launch height.

vx = v0x

vy = –v0y

Projectile Motion Analyzed


Further examining the equation,20

0 02 sin cosvRg

• If the launch height and the landing height are the same, then the maximum horizontal range is achieved if the launch angle is 45°.

0 0 0sin 2 2sin cos , Using the identity

we obtain20

0sin 2v

Rg

R is maximum when sin2θ0 = 1 or θ0 =45°.




• The launch height and the landing height differ.

• The launch angle 45° does not yield the maximum horizontal distance.


Projectile Motion AnalyzedThe Effects of the Air

Path I: Projectile movement if the air resistance is taken into account

Path II: Projectile movement if the air resistance is neglected (as in a vacuum)Our calculation along this chapter is based on this assumption


30

A pitcher throws a baseball at speed 40 km/h and at angle θ = 30°.

(a) Determine the maximum height h of the baseball above the ground.

h

0y yv v gt

0 01 2

0 0 2

02 2

0 02 ( )

x

y

y y

y y

x x v ty y v t gtv v gt

v v g y y

0 5.56 9.8t 5.569.8

t 0.567 s1 2

0 0 2yh y y v t gt 1 22(5.56)(0.567) (9.8)(0.567)

0 0 sin(11.11)sin 305.56 m s

yv v

0 0 cos(11.11)cos309.62 m s

xv v

40km h 11.11 m s

1.58 m

Example: Baseball Pitcher


30

A pitcher throws a baseball at speed 40 km/h and at angle θ = 30°.

d

(c) Determine the horizontal distance d it travels.

on air up downt t t 0.567 0.567 1.134 s (b) Determine the duration when the baseball is on the air.

0 01 2

0 0 2

02 2

0 02 ( )

x

y

y y

y y


v v g y y

0 0xd x x v t (9.62)(1.134)10.91 m

Example: Baseball Pitcher


Released horizontallyA rescue plane flies at 198 km/h and constant height h = 500 m toward a point directly over a victim, where a rescue capsule is to land.(a) What should be the angle Φ

of the pilot’s line of sight to the victim when the capsule release is made?

0h y y

1 22( 500) (0) (0) (9.8)t t

1 20 0 2yy y v t gt

2 2 500 102.0419.8

t

10.102 st

0d x x

0 0xx x v t (55)(10.102) 555.61 m

1tan dh

1 555.61tan500

48.016

Example: Rescue Plane


(b) As the capsule reaches the water, what is its velocity v in unit-vector notation and in magnitude-angle notation?

→

Released horizontallyA rescue plane flies at 198 km/h and constant height h = 500 m toward a point directly over a victim, where a rescue capsule is to land.

0h y y

0d x x 0y yv v gt

(0) (9.8)(10.102)yv 99 m s

0x xv v55 m s

ˆ ˆ55i 99 j m sv

Unit-vector notation

113.252 m s 60.945v

Magnitude-angle notation

Example: Rescue Plane


A stuntman plans a spectacular jump from a higher building to a lower one, as can be observed in the next figure. Can he make the jump and safely reach the lower building?

0 01 2

0 0 2

02 2

0 02 ( )

x

y

y y

y y


v v g y y

1 20 0 2yy y v t gt

1 22( 4.8) (0) (0) (9.8)t t

2 2 4.8 0.989.8

t

0.99 st

0 0xx x v t (4.5)(0.99)4.46 m

Time for the stuntman to fall 4.8 m

Horizontal distance jumped by the stuntman in 0.99 s

He cannot make the jump

Example: Clever Stuntman


Homework 5: Three Point Throw

A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket. If he shoots the ball at a 40.0° angle with the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m.

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