AP Physics Kinematics Pretest
1.
Which of the following best represents a graph of horizontal position vs time for balls launched
horizontally at three different velocities from the same height?
2.
Which of the
following graphs best represent a graph of vertical position vs time for
balls launched horizontally at three different velocities from the same height?
3.
Which of the following graphs best represents the
horizontal velocities vs time for
balls launched horizontally at three different velocities from the same height?
4.
Which of the following graphs best represents the vertical acceleration vs time of ball
s launched
horizontally with three different velocities from the same height?
5.
Sketch a graph of the vertical velocity vs time of balls launched horizontally vs
time at three different velocitie
s from the same height?
6.
Sketch a graph of the horizontal acceleration vs time of balls launched horizontally vs
time at three different velocities from the same height?
0m/s
2
7.
A car decelerat
es. Which of the following vector diagrams could be true?
Slope =

9.8 m/s/s
Physics
–
Kinematics in 2 Dimensions.
v
s
=
d
v =
x
v =
x
t
0
t
t
t
a =
v
–
v
o
t

t
o
v = v
fx
+ v
ix
v
f
=v
ix
+ at
x = x
o
+ v
ix
t + 1/2 at
2
v
f
2
= v
i
2
+ 2 a
x
2
x = v
ix
t + 1/2 at
2
v = v
f
+ v
iy
v=v
iy
+ gt
y = y
i
+ v
iy
t + 1/2 gt
2
v
y2
= v
iy
2
+ 2 g
y
2
y = v
iy
t + 1/2 gt
2
v
ix
= v
o
cos
v
iy
= v
o
sin
1.
Which
of the following best represents a graph of horizontal position as a function of time for
balls launched horizontally in the positive x direction?
t
2.
Which of the following best represents a graph of vertical position as a f
unction of time for balls
launched horizontally?
t
A
B
C
D
E
F
G
A
B
C
D
E
F
G
x
y
3.
Which of the following graphs best represents the horizontal velocities vs time for
balls launched horizontally?
time
4.
Which of the
following graphs best represents the vertical velocities vs time for
balls launched horizontally?
time
5.
Which of the following graphs best represents the horizontal acceleration vs time for
balls launched horizontally?
time
6.
Which of the following graphs best represents the vertical acceleration vs time for
balls launched horizontally?
A
B
C
D
E
F
G
A
B
C
D
E
F
G
0 m/s
0 m/s
A
B
C
D
E
F
G
0 m/s
2
0 m/s
2
A
B
C
D
E
F
G
7.
An Olympic athlete throws a javelin a
t four different angles above the horizon
tal, each with the
same speed: 2
0
o
3
0
o
5
0
o
7
0
o
Which two throws cause the javelin to land the same distance
away?
a)
20
o
and 70
o
b)
3
0
o
and 7
0
o
c)
2
0
o
and 5
0
o
d)
5
0
o
and 7
0
o
8.
Why
? Complement
ary angles create the same range. 20
0
+70
0
= 90
0
9. A projectile is fired at 30 degrees above the horizon. Which statement is true of the
y
–
component of its velocity
a) it is greater than the x component of its velocity
b) it
is less than the x component of its velocity
c) it is equal to the x component of its velocity
d) its y component of its velocity is

9.8 m/s
2
10
.
Why

less than 45
o
will create a greater horizontal (x) velocity
11. Which of the f
ollowing will have the greatest time of flight when fired from the
ground to the ground?
a) A projectile fired at 100 m/s at 25
o
degrees
b) A projectile fired at 100 m/s at 35
o
degrees
c) A projectile fired at 100 m/s at 45
o
degrees
d)
A project
ile fired at 100 m/s at 55
o
degrees
12.
Why?
The larger the angle the greater the time of flight
13. Which of the following will travel the greatest horizontal distance when fired from the
ground to the ground?
a) A projectile fir
ed at 100 m/s at 10
o
degrees
b) A projectile fired at 100 m/s at 20
o
degrees
c)
A projectile fired at 100 m/s at 50
o
degrees
d) A projectile fired at 100 m/s at 70
o
degrees
e) A projectile fired at 100 m/s at 80
o
degrees
14. Why?
The closer
the angle is to 45
o
the greater the horizontal distance.
Optimal time of flight and horizontal velocity occurs at 45
o
15. A projectile is fired at 45 degrees above the horizon. Which statement is true of the
y
–
component of its acceleration
a) the y
–
component of its acceleration at its maximum height is 0 m/s
2
b) the y
–
component of its acceleration initially is 9.8 m/s
2
sin 45
0
c) the y
–
component of its acceleration i
nitially is

9.8 m/s
2
sin 45
0
d
)
the y
–
component of its acceleration at its maximum height is

9.8 m/s
2
16. Why
? The y component of a projectile is always

9.8m/s/s directed down.
17 . You are throwing a ball for the second time.
If the ball leaves your hand with twice the velocity
it had on your first throw, its time of flight will be
a.
1.4 times as much.
b.
half as much
. v = v
o
+ at
c.
twice as much.
–
v
o
= t
d. four times as
much a
e. the same
18. You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity
it had on your first throw, the maximum height it will achieve is
a.
1.4 times as much
.
V
2
=
v
o
2
+ 2 g
y
b. half as much.
v
o
2
= 2 g
y
c. twice as much.
d.
four times as much
e. the same
19. You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity
it had
on your first throw, the vertical veloicity at the maximum height will be
a. 1.4 times as much.
b. half as much.
c. twice as much.
d. four times as much
e
. the same
20.
Projectiles that are fired down at 10 deg
rees below the horizon will have
a) a higher y velocity when they hit the ground when compared to a projectile fired
up at 10 degrees above the horizon at the same speed
b) a lower y velocity when they hit the ground when compared to a p
rojectile fired
up at 10 degrees above the horizon at the same speed
c)
the same y velocity when they hit the ground when compared to a projectile fired
up at 10 degrees above the horizon at the same speed
.
d) have a velocity that c
ould be higher or lower when compared to a projectile fired
up at 10 degrees above the horizon depending on the whether the initial velocities
are above 10 m/s or below 10 m/s.
Dropped
Fired at 10 m/s
1.20 m
1.20 m
21. A ball is dropped 1.20 m. A second ball is fired horizontally at 10 m/s.
a) What is the time of flight of the dropped ball?
b) What is the time of flight for the ball fired at 10 m/s
horizontally?
c) What is the horizontal distance of the ball fired at 10 m/s?
d) What is the horizontal and vertical positions of the ball fired 10 m/s horizontally .25 seconds
into its flight?
a
) Given
Find
F/R/S
礽

ㄮ㈰1†††††††⁴㴿=
礽y
oy
t + ½ g t
2
g=

9.8m/s
2
y=1/gt
2
2
礠
㴠=‽†
㈠ㄮ㈰2
= .495s
g 9.8m/s/s
b) Given
Find
F/R/S
礽

ㄮ㈰1†††††††⁴㴿=
礽癯祴₽⁴
2
g=

9.8m/s
2
y=1/gt
2
2
礠
㴠=
㴠=
㈠ㄮ㈰2
= .495s
g 9.8m/s/s
c)
x=x
o
+ v
ox
t + ½ a
t
2
x=x
o
+ v
ox
t + ½ a
t
2
x=
v
ox
t
10m/s (.495s) = 4.95 m
d)
x=x
o
+ v
ox
t + ½ a
t
2
x=x
o
+ v
ox
t + ½ a
t
2
x=
v
ox
t
10m/s (.25s ) = 2.50
m
y=y
o
+ v
oy
t + ½ g t
2
y=1.20m
+ ½ (

9.8 m/s
2
) (.25s)
2
= 0.894 m
18.
A ball is launched from .25 m off the ground at angle of 65
o
above the horizon at 10.0 m/s and lands on
the table .98 m high.
a)
Determine the initial horizontal velocity.
10m/s cos 65
o
= 4.23 m/s
b)
Determine the initial vertical velocity
10 m/s sin 65
o
= 9
.06 m/s
c)
Determine the time it takes the ball to reach its maximum height
v
y
=0m/s v = v
oy
+ g t

v
oy
= t =

9.06 m/s
= .924 s
g

9.8 m/s/s
d)
Determine the maximum height achieved by the ball.
y
=y
o
+ v
oy
t + ½ g t
2
y=.25 m + 9.06 m/s( .924 s ) + ½ (

9.8m/s/s) (.924s)
2
= 4.33 m
e)
Determine the horizontal position of the ball at the maximum height.
v
x
=d v
x
t = d 4.23m/s (.924s) = 3.91 m
t
x=x
o
+ v
ox
t + ½ a
t
2
x=x
o
+ v
ox
t + ½ a
t
2
x=
v
ox
t
4.23 m/s (.924s) = 3.91
m
f)
Determine the total time of flight from the .25 m vertical launch position to the .98 m vertical
landing position.
y=y
o
+ v
oy
t + ½ g
t
2
.98m=.25m + (9.06 m/s ) t + ½ (

9.8m/s/s) t
2
0 m =

.73 m + (9.06 m/s ) t + ½ (

9.8m/s/s) t
2
C=

.73
B= +9.06 A =

4.9
t=1.76s
g)
Determine the horizontal landing position of the ball.
v
x
=d v
x
t = d 4.23m/s (1.76s) = 7.44m m
t
x=x
o
+ v
ox
t + ½ a
t
2
x=x
o
+ v
ox
t + ½ a
t
2
x=
v
ox
t
4.23 m/s (1.76 s) = 7.44
m
h)
Determine the angle of the ball as it lands in the can.
Givens
Find
FR
S
t=1.76s
v
y
v
y
=v
oy
+ gt = 9.06m/s +

9.8m/s
2
(1.76s)=

8.18 m/s
inv tan (

8.18m/s / 4.23 m/s ) = 62.7
0
from the horizontal
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