Bulldog Physics
2D Kinematics
Texas Lutheran University
Example Problem 1: 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?
Separate out x- and y- components:
x
v0 x v0 cos(35 ) (20.1 ms ) cos(35 ) 16.5 ms
ax
0
v0 y
y
v0 sin(35 ) (20.1 ms ) sin(35 ) 11.5 ms
m
s2
9.81 sm2
ay
x ?
t ?
y
t
0m
?
(a) First, find t:
v0 y t
1 2
a yt
2
1 2
ayt
2
t
y
0
v0 y t
2(11.5 ms )
9.81 sm2
t
2v0 y
ay
2.35s .
(b) Now use t to find x:
x
x v0 xt (16.5 ms )(2.35s) 38.8m .
t
(c) For finding the maximum height, our table changes a bit:
y
v0 y v0 sin(35 ) (20.1 ms ) sin(35 ) 11.5 ms
v0 x
0 ms
v fy
9.81 sm2
ay
y
?
The maximum height is found by using
v 2fy
v02y
y
Page 1
2a y ( y ) 2
y
(0 ms ) 2 (16.5 ms ) 2
2( 9.81 sm2 )
v 2fy v02y
2a y
3.73m
Department of Physics
Texas Lutheran University
Seguin, TX 78155
Bulldog Physics
2D Kinematics
Texas Lutheran University
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?
Separate out x- and y- components:
x
v0 x ?
ax
0
v0 y
m
s2
0
m
s
y
(horizontal throw)
ay
x 38.0m
t ?
y
9.81 sm2
1.83m (displacement is down)
t ?
First, find t:
y
y
1 2
a yt
2
v0 y t
1 2
a yt
2
t
2 y
ay
t
2( 1.83m)
9.81 sm2
0.611s .
v0 x
x
t
62.2 ms .
Now use t to find v0:
Page 2
38.0m
0.611s
Department of Physics
Texas Lutheran University
Seguin, TX 78155
Bulldog Physics
2D Kinematics
Texas Lutheran University
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?
(a) Separate out x- and y- components:
x
v0 x v0 cos(44 ) (24.0 ms ) cos(44 ) 17.3 ms
ax
0
v0 y
y
v0 sin(44 ) (24.0 ms ) sin(44 ) 16.7 ms
m
s2
9.81 sm2
ay
x ?
t ?
y
15.2m (displacement is down)
t ?
(a) First, find t:
y
1 2
ayt
2
v0 y t
1 2
a yt
2
v0 y t
y
0
Use the quadratic equation
t
v0 y
( v0 y ) 2
4( 12 a y )(
y)
16.7 ms
( 16.7 ms ) 2 2( 9.81 sm2 )(15.2m)
2( 12 a y )
9.81 sm2
t { 0.747s, 4.15s}
The positive time is the physical solution so t 4.15 seconds. Now to find x:
x
v0 x
x v0 xt (17.3 ms )(4.15s) 71.8m .
t
(b) Now we need to find vfx and vfy. The velocity in the x-direction is constant (no acceleration)
so v fx 17.3 ms . To find vfy:
v fy
v0 y
at
(16.7 ms ) ( 9.81 sm2 )(4.11s)
23.6 ms .
The final speed is found using the Pythagorean Theorem:
vf
v 2fx v 2fy
(17.3 ms )2 ( 23.6 ms )2
29.3 ms .
The direction of the velocity below the horizontal is found by using tangent:
v fy
23.6 ms
tan
tan 1
53.8 .
v fx
17.3 ms
Page 3
Department of Physics
Texas Lutheran University
Seguin, TX 78155