Transcript Document 7612028

Chapter 4
Motion in Two Dimensions
EXAMPLES
Example 4.1 Driving off a cliff.
yi = 0 at top, y is positive upward.
Also vyi = 0
 How fast must the motorcycle
leave the cliff to land at
x = 90 m, y = – 50 m
 Unknown: vxi = ?
 Formulas: vy = gt
x = vxit y = ½gt2
 Time to Bottom:

t

2  50m 
2y

 3.19s
g
9.8m / s 2
vxi = x/t = 90.0m/3.19s 
vxi = 28.2 m/s
Example 4.2 Kicked football
vyi
vi
vxi


Given: θi = 37º, vi = 20 m/s
vxi = vicosθi = 16 m/s & vyi= visinθi = 12 m/s
Find
a. Max height (h) ? b. Time when hits ground?
c. Total distance traveled in the x direction (R) ?
d. Velocity at top?
e. Acceleration at top?
Example 4.2 cont.
Using Eq: 4.13
(a)
(b)
(c)
(d )
(e)
(vi sin i )2
( 12m/s)2
144m2 / s 2
h


 7.35m
2
2
2g
2( 9.8m/s ) 19.6m / s
At t  0, y  0. At the end when the ball touches the ground:
y  0  Using Eq. 4.12:
f
y  (vi sin θi )t  1 gt 2  0  (12m / s)t  4.9(m / s 2 )t 2  Solving for t:
2
12m / s
t  0 (of course!) and t 
 2.45s
2
4.9m / s
R  v  t  (v cos θ )t  ( 16m/s)( 2.45s)  39.2m
xi
i
i
At this point: v  0  v  v  16m / s
y
x
xi
The acceleration vector is the same at the highest
point as it is throughout the flight, which is:  9.80 m/s 2
Example 4.3
Where Does The Apple Land?

A child sits in a wagon, moving to the
right (x-direction) at constant velocity
vox. She throws an apple straight up
(from her viewpoint) with initial velocity
voy while she continues to travel forward
at vox Neglect air resistance.

Will the apple land behind the wagon,
in front of the wagon, or in the wagon?
Example 4.3 Cont.



The apple will stay above the
girl the entire trip and will
land in the wagon.
The reason is:
To a person in the ground
reference frame (b) the apple
will be exactly a projectile in
motion (neglecting air
resistance). To the girl it is an
object in free fall.
And the Vertical motion of a
projectile and free fall are
the same.
Example 4.4 Wrong Strategy
(Similar to Example 4.3 Text Book)
“Shooting the Monkey”!!
 A boy on a small hill aims his
water-balloon slingshot
horizontally, straight at a second
boy hanging from a tree branch a
distance d away. At the instant the
water balloon is released, the
second boy lets go an fall from the
tree, hopping to avoid being hit.
 Show that he made the wrong
move (He hadn’t studied
Physics yet!!)
Example 4.4 Cont.
“Shooting the Monkey”!!



Both the water balloon and the boy in the tree start falling at
the same time, and in a time t they each fall the same vertical
distance y = ½gt2
In the same time it takes the water balloon to travel the
horizontal distance d, the balloon will have the same y
position as the falling boy.
Splash!!! If the boy had stayed in the tree, he would have
avoided the humiliation
Example 4.5 That’s Quite an Arm

Non-Symmetric Projectile
Motion

Example 4.4 (text book), page 84
Follow the general rules for projectile
motion
Break the y-direction into parts
 up and down or
 symmetrical back to initial height and
then the rest of the height
May be non-symmetric in other ways



Example 4.5 Cont.






Given: θi = 30º, vi = 20 m/s 
(A) vxi= vicosθi = 17.3 m/s and vyi= visinθi = 10.0 m/s
At t = 0 : xi = 0 yi = 0
Find: t = ? (time at which the stone hits the ground) with yf = –
45.0m
Using: yf = vyit – ½gt2 
– 45.0m = (10.0m/s)t – (4.90m/s2)t2
Solving for t using General Quadratic Formula: t = 4.22 s
(B) vxi= vxf = 17.3 m/s and
vyf = vyi – gt 
vyf = 10.0m/s – (9.80m/s2)(4.22s) 
vyf = ̶ 31.4m/s 
 v 2f  vxf2  v yf2  v f  vxf2  v yf2
v f  (17.3) 2  (31.4) 2 m / s  35.9m / s
Example 4.6 The End of the Ski Jump
Example 4.5 (text book), page 85
 Given the figure of the ski jumper, find
the distance d traveled along the incline.
1. Coordinates x and y at the end:

x f  v xi t  25t
(1)
y f   12 gt 2  4.9t 2
2. From the figure:
x f  d cos 35 (3)
y f  d sin 35
(4)
(2)
Example 4.6 Cont.
3. Equating (1) = (3) and (2) = (4)
d cos 35  25t
d sin 35  4.9t 2
(5)
(6)
4. Dividing (6) by (5):
tan 35 
4.9
t
25
 t  3.57 sec (7)
5. Substitution of (7) in (5) and solving for d:
d cos 35  25(3.57)  d  109m
6. Substitution of d into (3) and (4), gives the coordinates:
x f  (109) cos 35  89.3m
y f  (109) sin 35  62.5m
Example 4.7
The Centripetal Acceleration of the Earth

Calculate ac of the Earth, assuming it moves in a circular
orbit around the Sun.
 2 r 
v 2  T  4 2 r 2 4 2 r
ac  

 2 
2
r
r
rT
T
2
4 2 1.496 x1011 m 
1yr

3
2
ac 

5.93
x
10
m
/
s

2
7 
3.156
x
10
s

1
yr
 
2


Note that ac << g

Material for the Midterm

Material from the book to Study!!!



Objective Questions: 1-4-6
Conceptual Questions: 5-6-7
Problems: 2-4-6-10-11-21-27