Two-Dimensional Kinematics
v

v

2
a
(
x

x
)
f
0
0
2
2
v

v

a
t
f
0
1 2
x
x

v
t a
t
0
0
2
Bulldog Physics
1
Texas Lutheran University
Department of Physics
Seguin, TX 78155
y
Projectile Motion
What is a Projectile?
x
Any object that flies through the air.
Projectile motion is a example of 2D Motion
An object flying through the air moves in both the x and y directions.
Typically, the motion in each component is uncoupled.
The motion in each component is independent of the other.
What is the acceleration of an object while it’s flying
through the air?
j
Acceleration due to gravity! agˆ
What is the acceleration in the x-direction?
It is ZERO!!!
ax 0
What does this mean about the velocity in the x-direction?
It is constant.
2
Texas Lutheran University
Department of Physics
Seguin, TX 78155
Conceptual Question 1
A blue ball is being held above ground a distance y.
A red ball is thrown from the same height, y, at a
horizontal velocity of v0. The blue ball is released at
the same instant the red ball leaves the table. Which
ball reaches the ground first?
v0
1) The blue ball lands first.
y
2) The red ball lands first.
3) They land simultaneously.
4) There is not enough information.
3
Texas Lutheran University
Department of Physics
Seguin, TX 78155
Conceptual Question 1
A blue ball is being held above ground a distance y.
A red ball is thrown from the same height, y, at a
horizontal velocity of v0. The blue ball is released at
the same instant the red ball leaves the table. Which
ball reaches the ground first?
v0
1) The blue ball lands first.
y
2) The red ball lands first.
3) They land simultaneously.
4) There is not enough information.
4
Texas Lutheran University
Department of Physics
Seguin, TX 78155
y
A Ball in the Air
We can separate the motion in the x and y directions.
x-direction
y-direction
ax  0
ay  g
v0,x  v0co
s
v0,y  v0sin
x
Now we use the kinematics
equations using only one
column at a time to solve
our problems.
y  0
t
t
We also can talk about the time in the air.
x  L
This must be the same for both directions!
So t can go in both columns.
What is the velocity at max height?
v0
ax  0 , so vx is constant.

So only vy is zero at the maximum height.
5
L
Texas Lutheran University
Department of Physics
Seguin, TX 78155
Example Problem 1
6
Long TD Pass video
Texas Lutheran University
Department of Physics
Seguin, TX 78155
Example Problem 1
Play “Long TD Pass”
A quarterback throws a football at 45 mph (20.1 m/s) at 35
degrees with the horizontal to get the ball over the linemen.
Assume the ball is caught at the same height it is thrown.
(a) How long is the ball in the air?
(b) How far away can the receiver be and catch the ball?
(c) What is the maximum height of the ball while in the air?
7
Texas Lutheran University
Department of Physics
Seguin, TX 78155
y
A Ball in the Air (Horizontal Velocity Case)
We still have the same basic columns.
x-direction
y-direction
ax  0
ay  g
v0,x  v 0
x
v0,y  0
y  h
t
t
We now have a change in height, so y is no longer zero.
Our initial velocity is now completely in the x-direction.
x  L
Again, we’re now set up to
solve problems using the
kinematics equations using
one column at a time.
v0
h
L
8
Texas Lutheran University
Department of Physics
Seguin, TX 78155
Example Problem 2
A thirdbaseman catches a ground ball and throws to the
firstbaseman, who is 38.0 meters away. The thirdbaseman
throws the ball horiztonally at height of 1.83 meters off
the ground. What is the minimum speed the thirdbaseman
can throw the ball without the ball hitting the ground
before reaching the firstbaseman?
9
Texas Lutheran University
Department of Physics
Seguin, TX 78155
A Ball in the Air (Angle and Height)
y
We still have the same basic columns.
x-direction
y-direction
ax  0
ay  g
x
1 2

y

vt
a
t
v0,x v0co
0
0
,y
y
s v0,y  v0sin y
2
g2
y  h
x  L

h

s
i
n

v
t t
0
2
t
t
When this situation arises, we almost always get a
quadratic equation when analyzing the y-direction.
v0
g
2
t
v
s
i
n

t
h

0


0

2
h
How do we solve this?
L
10
Texas Lutheran University
Department of Physics
Seguin, TX 78155
The Quadratic Equation
We use an algebra trick:
If we have an equation of the form:
g
2
t
v
s
i
n

t
h

0


0
2
A
t
B
tC


0
2

B B
4
A
C
Then the solution is:
t
g
2
A
A
For the situation at hand, we have
2
B


v
s
i
n

0
g
2
2
v
s
in
 v0sin4 
h

0
2
 
C


h
t
So we get
2
g
2
 
2


v
s
i
nv
s
i
n
2
g
h
0
0
t
g
11
2
2
The plus and minus give you up
to two possible solutions. This
means there are two times
where you reach the specified
height.
Texas Lutheran University
Department of Physics
Seguin, TX 78155
Example Problem 3
A golfer tees off on a par 3 course. The she lofts the ball at a 44
degree angle from the horizontal, sending it off at 24 m/s. The hole
is positioned 50 feet (15.2 meters) below the tee. The ball lands
right into the hole!
(a) What is the horizontal distance between the tee and the hole?
(b) How fast the ball traveling just before it lands in the hole?
12
Texas Lutheran University
Department of Physics
Seguin, TX 78155
Conceptual Question 2
A quarterback throws two
passes, one to Receiver A
and the second to Receiver
B. The trajectories are
shown. Which pass takes
more time to get to the
receiver?
1) The pass to Receiver A.
2) The pass to Receiver B.
3) Both passes spend the same time in the air.
4) There is not enough information.
13
Texas Lutheran University
Department of Physics
Seguin, TX 78155
Conceptual Question 2
A quarterback throws two
passes, one to Receiver A
and the second to Receiver
B. The trajectories are
shown. Which pass takes
more time to get to the
receiver?
1) The pass to Receiver A.
Receiver A
Receiver B
2) The pass to Receiver B.
3) Both passes spend the same time in the air.
4) There is not enough information.
14
Texas Lutheran University
Department of Physics
Seguin, TX 78155