Kinematics/ Projectile review
Problem
1.
An arrow is shot vertically upward with an initial speed of 25 m/s. When it’s exactly halfway to the top of its
flight, a second arrow is launched vertically upward from the same spot. The second arrow reaches th
e first
arrow just as the first arrow reaches its highest point.
(a) What is the launch speed of the second arrow?
(b) What maximum height does the second arrow reach?
2.
Two planes fly from Toronto to Philadelphia. Plane A flies via Pittsburgh whereas p
assengers on plane B
have a direct flight. Pittsburgh is 350 km due south of Toronto and 390 km due west of Philadelphia. The
airspeed of both planes is 400.0 km/h and a steady wind is blowing from the east at 60.0 km/h.
(a) What direction must the pilot p
oint the plane flying from Toronto to Pittsburgh? Include a vector diagram
of velocities.
(b) How long will the entire flight take for plane A assuming a 0.50

h layover in Pittsburgh?
(c) How much time must the pilot of plane B wait before leaving Toronto
if she is to arrive in Philadelphia at
the same time plane A arrives?
3.
The graph below represents the motion of an object over a recorded time interval. Using methods of graphical
analysis wherever possible, determine
(a) the object’s displacement rela
tive to its starting position at
t
= 6.0 s.
(b) the object’s average velocity between
t
= 0.0 s and
t
= 6.0 s.
(c) the object’s average speed between
t
= 0.0 s and
t
= 6.0 s.
(d) Including
t
= 0.0 s, how many times during the entire recorded time interval
is the object at its starting
position?
(e) During which interval is the object’s acceleration the greatest? What is the value of the acceleration during
this interval?
(f) Plot the corresponding position

time graph.
(g) Plot the corresponding acceleration

time graph.
4.
A football quarterback attempts a pass to one of the receivers. As the ball is snapped, the receiver leaves the
line of scrimmage and runs directly down field. The quarterback releases the ball 2.0 s later and from a
position 3.0 m behi
nd the line of scrimmage. He throws the ball with a speed of 26 m/s at an elevation of 60
above the horizontal. The receiver makes a diving reception, catching the ball just as it reaches the ground.
See the diagram below.
(a) What is the time of flight
of the football?
(b) What is the average speed of the receiver?
5.
A circus clown is fired from a cannon into a net that is situated 2.0 m above the cannon and some distance
from it. The cannon is elevated at 50.0
to the horizontal and the clown’s spee
d at launch is 15 m/s. See the
diagram below.
(a) Find the horizontal distance from the cannon where the net needs to placed in order for the clown to land
in it.
(b) Calculate the clown’s velocity as he lands in the net.
Kinematics/ Projectile review
A
nswer Section
PROBLEM
1.
ANS:
(a)
Using the sign convention that “up” is (
–
) and “down” is (+):
v
1
=
–
25 m/s
v
2
= 0.0 m/s
a
= 9.8 m/s
2
d
= ?
The arrow travels 31.9 m upward to its highest point. The halfway position is 15.9 m.
The time to travel the last half of its flight:
d
=
–
15.9 m
v
2
= 0.0 m/s
a
= 9.8 m/s
2
t
= ?
For the second arrow:
d
=

31.9 m
a
= 9.8 m/s
2
t
= 1.80 s
v
1
= ?
The speed of the second arrow at launch is 27 m/s [upward].
(b)
Finding the maximum height of the second arrow:
v
1
=
–
26.5 m/s
v
2
= 0.0 m/s
a
= 9.8 m/s
2
d
= ?
The second arrow reaches a maximum height of 36 m [upward].
REF:
K/U
OBJ:
1.3
LOC:
FM1.02
2.
ANS:
(a) The triangle of velocity vectors appears as:
The plane must point [8.6
f⁓].
(b) For the first leg of the trip of plane A:
time to fly from Toronto to Pittsburgh:
Layover time in Pittsburgh is 0.5 h.
Pittsburgh to P
hiladelphia:
Time to fly:
Total time Toronto to Philadelphia: 0.885 h + 0.5 h + 1.15 h = 2.5 h
The total time for plane A is 2.5 h.
(c) Distance from Toronto to Philadelphia:
Vector triangle of velocities:
Using sine law:
= 5.8
, then
= 18
0
–
138
–
5.8
= 36
Using cosine law:
353 km/h
Time for plane B to fly from Toronto to Philadelphia:
Plane B must wait 2.5 h
–
1.48 h = 1.0 h.
REF:
K/U
OBJ:
1.5
LOC:
FM1.05
3.
ANS:
(a)
displacement
= area under graph
= 23.75 m [S] + 18.75
m [N]
displacement
= 5.0 m [S]
(b)
The object’s average velocity during the first 6.0 s is 0.83 m/s [S].
(c)
The object’s average speed during the first 6.0 s is 7.1 m/s.
(d) The object is at its starting location 3 times throughout the motion.
(e) The object’s acceleration is greatest between
t
= 6.5 s and 7.0 s. (the greatest slope) acceleration = slope of
graph = 30 m/s
2
[N]
(f)
(g)
REF:
K/U
OBJ:
1.2
LOC:
FM1.02
4.
ANS:
(a)
Time of flight: let “up” be (
–
) and “down” be (+)
v
1
=
–
26 m
/s(sin 60º) =
–
22.5 m/s
a
= 9.8 m/s
2
d
= 2.0 m
t
= ?
2.0 = (
–
22.5)
t
+ 4.9(
t
)
2
Solving the quadratic:
t
= 4.68 s
The time of flight is 4.7 s.
(b)
Horizontal range:
d
=
v
t
= 26 m/s(cos 60
)(4.68 s) = 60.8 m
The receiver must run: 60.8 m
–
3.0 m =
57.8 m.
The time the receiver has to reach the football: 4.68 s + 2.0 s = 6.68 s.
The average speed of the receiver:
The receiver must run with an average speed of 8.7 m/s.
REF:
K/U
OBJ:
1.4
LOC:
FM1.03
5.
ANS:
(a)
Time of flight: let “up” be (
–
) an
d “down” be (+)
v
1
=
–
15 m/s(sin 50
) =
–
11.5 m/s
a
= 9.8 m/s
2
d
=
–
2.0 m
t
= ?
–
2.0 = (
–
11.5)
t
+ 4.9(
t
)
2
Solving the quadratic:
t
= 0.19 s (way up) and 2.16 s (way down)
Horizontal range:
d
=
v
t
= 15 m/s(cos 50º)(2.16 s) = 21 m
The net must be placed 21 m away from the cannon.
(b) Horizontal component of final velocity: 15 m/s(cos 50
) = 9.64 m/s
Vertical component of final velocity:
v
2
=
v
1
+
a
t
=
–
11.5 m/s + 9.8 m/s
2
(2.16 s)
v
2
= 9.67 m/s
Using Pythagoras:
=
The shell lands with a velocity of 14 m/s at an angle of 45
扥bow t桥h物z潮o慬.
REF:
K/U
OBJ:
1.4
LOC:
FM1.03
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