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University Physics: Mechanics. Ch 4 . TWO- AND THREE-DIMENSIONAL MOTION. Lecture 5. Dr.-Ing. Erwin Sitompul. http://zitompul.wordpress.com. Homework 4: The Plane. A 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. - PowerPoint PPT Presentation
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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.
5/3Erwin Sitompul University Physics: Mechanics
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→
5/4Erwin Sitompul University Physics: Mechanics
(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
5/5Erwin Sitompul University Physics: Mechanics
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 ˆi j kyx zdvdv dv
dt dt dt
Average and Instantaneous Acceleration→ →
→
→→
5/6Erwin Sitompul University Physics: Mechanics
We can rewrite the last equation asˆ ˆ ˆi 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
→
5/7Erwin Sitompul University Physics: Mechanics
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,
5/8Erwin Sitompul University Physics: Mechanics
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.
→
5/9Erwin Sitompul University Physics: Mechanics
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
5/10Erwin Sitompul University Physics: Mechanics
Projectile Motion
0xa
ya g0 0 0
ˆ ˆi jx yv v v
0 0 0cosxv v
0 0 0sinyv v
5/11Erwin Sitompul University Physics: Mechanics
Projectile MotionTwo Golf Balls
• The vertical motions are quasi-identical.
• The horizontal motions are different.
0 0xv
0 0xv
5/12Erwin Sitompul University Physics: Mechanics
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
5/13Erwin Sitompul University Physics: Mechanics
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
5/14Erwin Sitompul University Physics: Mechanics
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°.
Projectile Motion Analyzed
5/15Erwin Sitompul University Physics: Mechanics
Projectile Motion Analyzed
• The launch height and the landing height differ.
• The launch angle 45° does not yield the maximum horizontal distance.
5/16Erwin Sitompul University Physics: Mechanics
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
5/17Erwin Sitompul University Physics: Mechanics
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
5/18Erwin Sitompul University Physics: Mechanics
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
x x v ty y v t gtv v gt
v v g y y
0 0xd x x v t (9.62)(1.134)10.91 m
Example: Baseball Pitcher
5/19Erwin Sitompul University Physics: Mechanics
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
5/20Erwin Sitompul University Physics: Mechanics
(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
5/21Erwin Sitompul University Physics: Mechanics
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
x x v ty y v t gtv v gt
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
5/22Erwin Sitompul University Physics: Mechanics
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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