When you throw
a football across the yard to your friend, you are using
physics. You make adjustments for all the factors, such as
distance, wind and the weight of the ball. The farther away
your friend is, the harder you have to throw the ball, or the
steeper the angle of your throw. This adjustment is done in
your head, and it's physics  you just don't call it that
because it comes so naturally.
Physics is the branch of science that deals with the
physical world. The branch of physics that is most relevant to
football is mechanics, the study of motion and its
causes. We will look at three broad categories of motion
as they apply to the game:
 Delivery of a football through the air
 Runners on the field
 Stopping runners on the field
Watching a weekend football game could be teaching you
something other than who threw the most passes or gained the
most yards. Football provides some great examples of the basic
concepts of physics  it's present in the flight of the ball,
the motion of the players and the force of the tackles. In
this edition of HowStuffWorks,
we'll look at how physics applies to the game of football.
Throwing the Football
The Language of
Physics
 Acceleration  Rate of change of velocity
with respect to time (calculated by subtracting the
starting velocity from the final velocity and dividing
the difference by the time required to reach that
final velocity)
 Force  influence on a body that causes it
to change speed or direction
 Velocity  Speed and direction that an
object travels (distance traveled divided by the
elapsed amount of time)
 Speed  How quickly an object moves
(distance traveled divided by the elapsed amount of
time)
 When
the football travels through the air, it always follows a
curved, or parabolic, path because the movement of the
ball in the vertical direction is influenced by the force of
gravity.
As the ball travels up, gravity slows it down until it stops
briefly at its peak height; the ball then comes down, and
gravity accelerates it until it hits the ground. This is the
path of any object that is launched or thrown (football,
arrow, ballistic missile) and is called projectile
motion. To learn about projectile motion as it applies to
football, let's examine a punt (Figure 1). When a punter kicks
a football, he can control three factors:
 The velocity or speed at which the ball leaves his foot
 The angle of the kick
 The rotation of the football
The rotation of the ball  spiral or endoverend  will
influence how the ball slows down in flight, because the ball
is affected by air drag. A spiraling kick will have less air
drag, will not slow down as much and will be able to stay in
the air longer and go farther than an endoverend kick. The
velocity of the ball and the angle of the kick are the major
factors that determine:
 How long the ball will remain in the air (hangtime)
 How high the ball will go
 How far the ball will go
The angle of a
kick helps determine how far it will
travel. 
Football by the
NumbersSince physics
is a quantitative science, developing some units and
measures is a good way to begin to understand the
effects of physics on football. Consider these useful
numbers and units developed by Dr. David Haase of North
Carolina State University:
 Player at full speed  ~22 miles per hour
(9.8 m/s)
 Linebacker  ~220 pounds (98 kg)
 Offensive lineman  ~300 pounds (133 kg)

When the ball leaves the punter's foot, it is moving with a
given velocity (speed plus angle of direction)
depending upon the force with which he kicks the ball.
The ball moves in two directions, horizontally and vertically.
Because the ball was launched at an angle, the velocity is
divided into two pieces: a horizontal component and a vertical
component. How fast the ball goes in the horizontal direction
and how fast the ball goes in the vertical direction depend
upon the angle of the kick. If the ball is kicked at a steep
angle, then it will have more velocity in the vertical
direction than in the horizontal direction  the ball will go
high, have a long hangtime, but travel a short distance. But
if the ball is kicked at a shallow angle, it will have more
velocity in the horizontal direction than in the vertical
direction  the ball will not go very high, will have a short
hangtime, but will travel a far distance. The punter must
decide on the best angle in view of his field position. These
same factors influence a pass or field goal. However, a field
goal kicker has a more difficult job because the ball often
reaches its peak height before it reaches the uprights.
If you are not interested in the details of calculating the
hangtime, peak height and range of a punt, click
here to skip the following page.
Punting: HangTime, Peak Height and
Range The parabolic path of a football can be
described by these two equations:
 y = V_{y}t  0.5gt^{2}
 x =V_{x}t
 y is the height at any time (t)
 V_{y} is the vertical component of the
football's initial velocity
 g is acceleration due to Earth's gravity, 9.8
m/s^{2}
 x is the horizontal distance of the ball at any
time (t)
 V_{x} is the horizontal component of
the football's initial velocity
To
calculate the hangtime (t_{total}), peak
height (y_{max}), and maximum range
(x_{max}) of a punt, you must know the initial
velocity (V) of the ball off the kicker's foot, and the
angle (theta) of the kick.
 The velocity must be broken into horizontal
(V_{x}) and vertical (V_{y})
components according to the following formulas:
 V_{x} = V cos(theta)
 V_{y} = V sin(theta)
 The hangtime (t_{total}) must be
determined by one of these two formulas:
 t_{total} = (2V_{y}/g)
 t_{total} = (0.204V_{y})
 Once you know the hangtime, you can calculate maximum
range (x_{max}):
 x_{max} = V_{x}
t_{total}
 You can calculate the time (t_{1/2}) at
which the ball is at its peak height:
 And you can calculate the peak height
(y_{max}), using one of these two formulas:
 y_{max} = v_{y}(t_{1/2}) 
1/2g(t_{1/2})^{2}
 y_{max} = v_{y}(t_{1/2}) 
0.49(t_{1/2})^{2}
For
example, a kick with a velocity of 90 ft/s (27.4 m/s) at an
angle of 30 degrees will have the following values:
 Vertical and horizontal components of velocity:
 V_{x} = V cos(theta) = (27.4 m/s) cos
(30 degrees) = (27.4 m/s) (0.0.87) = 23.7 m/s
 V_{y} = V sin(theta) = (27.4 m/s) sin
(30 degrees) = (27.4 m/s) (0.5) = 13.7 m/s
 Hangtime:
 t_{total} = (0.204V_{y}) =
{0.204 (13.7m/s)} = 2.80 s.
 Maximum range:
 x_{max} = V_{x}
t_{total} = (23.7 m/s)(2.80 s) = 66.4 m
 1 m = 1.09 yd
 x_{max} = 72 yd
 Time at peak height:
 t_{1/2} = 0.5 t_{total} =
(0.5)(2.80 s) = 1.40 s
 Peak height:
 y_{max} = V_{y}(t_{1/2}) 
0.49(t_{1/2})^{2} = [{(13.7 m/s)(1.40
s)}  {0.49(1.40 s)^{2}}] = 18.2 m
 1 m = 3.28 ft
 y_{max} = 59.7 ft
If we
do the calculations for a punt with the same velocity, but an
angle of 45 degrees, then we get a hangtime of 3.96 s, a
maximum range of 76.8 m (84 yd), and a peak height of 36.5 m
(120 ft). If we change the angle of the kick to 60 degrees, we
get a hangtime of 4.84 s, a maximum range of 66.3 m (72 yd),
and a peak height of 54.5 m (179 ft). Notice that as the angle
of the kick gets steeper, the ball hangs longer in the air and
goes higher. Also, as the angle of the kick is increased, the
distance traveled by the ball increases to a maximum (achieved
at 45 degrees) and then decreases.
Runners on the Field When we look at runners
on the field, several aspects can be considered:
 Where they line up for a play
 Changing directions
 Running in an open field
LineUp Positions When we look at the
positions of the backs, both offensive and defensive, we see
that they typically line up away from the line of scrimmage on
either side of the offensive and defensive linemen. Their
positioning allows them room, or time, to accelerate
from a state of rest and reach a high speed, to either run
with the ball or pursue the ball carrier. Notice that the
linebackers have far more room to accelerate than the linemen,
and the wide receivers have far more room than the
linebackers. So linebackers can reach higher speeds than
linemen, and wide receivers can reach the highest speeds of
all.
Changing Directions on the Field
Let's look at an example of a running play in which
the quarterback hands the ball off to a running back. First,
the running back starts from the set position, at rest, and
accelerates to full speed (22 mi/h or 9.8 m/s) in 2 s after
receiving the ball. His acceleration (a) is:
 a = (v_{f}  v_{o})/(t_{f} 
t_{o})
 v_{f} is final velocity
 v_{o} is initial velocity
 t_{f} is final time
 t_{o} is initial time
 a=(9.8 m/s  0 m/s)/(2 s  0 s)
 a= 4.9 m/s^{2}
As he runs with
the flow of the play (e.g. to the right), he maintains
constant speed (a = 0). When he sees an opening in the line,
he plants his foot to stop his motion to the right, changes
direction and accelerates upfield into the open. By planting
his foot, he applies force to the turf. The force he
applies to the turf helps to accomplish two things:
 Stop his motion to the right
 Accelerate him upfield
To stop his motion to the
right, two forces work together. First, there is the force
that he himself applies to the turf when he plants his foot.
The second force is the friction
between his foot and the turf. Friction is an extremely
important factor in runners changing direction. If you have
ever seen a football game played in the rain, you have seen
what happens to runners when there is little friction to
utilize. The following is what happens when a runner tries to
change his direction of motion on a wet surface:
 As he plants his foot to slow his motion, the coefficient
of friction between the turf and him is reduced by the
water on the surface.
 The reduced coefficient of friction decreases the
frictional force.
 The decreased frictional force makes it harder for him
to stop motion his to the right.
 The runner loses his footing and falls.
The
applied force and the frictional force together must stop the
motion to the right. Let's assume that he stops in 0.5 s. His
acceleration must be:
 a = (0 m/s  9.8 m/s)/(0.5 s  0 s)
 a = 19.6 m/s^{2}
*The negative sign indicates that the runner is
accelerating is in the opposite direction, to the left.
The Language of
Physics
 Mass  The amount of substance that an
object contains
 Momentum  The mathematical product of the
mass of a moving object and its velocity
 Impulse  The mathematical product of force
and the time over which that force is applied to an
object

The force (F) required to stop him is the product of his
mass (m), estimated at 98 kg (220 lbs), and his acceleration:
 F = ma = (98 kg)(19.6 m/s^{2}) = 1921
Newtons (N)
 4.4 N = 1 lb
 F = ~500 lbs!
To accelerate upfield, he
pushes against the turf and the turf applies an equal and
opposite force on him, thereby propelling him upfield. This is
an example of Isaac Newton's third law of motion, which
states that "for every action there is an equal, but opposite
reaction." Again, if he accelerates to full speed in 0.5 s,
then the turf applies 1921 N, or about 500 lbs, of force. If
no one opposes his motion upfield, he will reach and maintain
maximum speed until he either scores or is tackled.
Running in an Open Field
When running in an open field, the player can reach
his maximum momentum. Because momentum is the product
of mass and velocity, it is possible for players of different
masses to have the same momentum. For example, our running
back would have the following momentum (p):
 p = mv = (98 kg)(9.8 m/s) = 960 kgm/s
For a 125 kg (275 lb) lineman to have the same
momentum, he would have to move with a speed of 7.7 m/s.
Momentum is important for stopping (tackling, blocking)
runners on the field.
Blocking and Tackling Tackling and blocking
runners relies on three important principles of physics:
 Impulse
 Conservation of momentum
 Rotational motion
Photo courtesy North
Carolina State University Players use physics to stop each other
on the football field.
When Runner and Tackler
Meet When our running back is moving in the open
field, he has a momentum of 960 kgm/s. To stop him 
change his momentum  a tackler must apply an impulse in the
opposite direction. Impulse is the product of the
applied force and the time over which that force is applied.
Because impulse is a product like momentum, the same impulse
can be applied if one varies either the force of impact or the
time of contact. If a defensive back wanted to tackle our
running back, he would have to apply an impulse of 960 kgm/s.
If the tackle occurred in 0.5 s, the force applied would be:
 F = impulse/t = (960 kgm/s)/(0.5 s) = 1921 N =
423 lb
Alternatively, if the defensive back
increased the time in contact with the running back, he could
use less force to stop him.
In any collision or tackle in which there is no force other
than that created by the collision itself, the total momentum
of those involved must be the same before and after the
collision  this is the conservation of momentum.
Let's look at three cases:
 The ball carrier has the same momentum as the tackler.
 The ball carrier has more momentum than the tackler.
 The ball carrier has less momentum than the tackler.
For the discussion, we will consider an elastic
collision, in which the players do not remain in contact
after they collide.
 If the ball carrier and tackler have equal momentum, the
forward momentum of the ball carrier is exactly matched by
the backward momentum of the tackler. The motion of the two
will stop at the point of contact.
 If the ball carrier has more momentum than the tackler,
he will knock the tackler back with a momentum that is equal
to the difference between the two players, and will likely
break the tackle. After breaking the tackle, the ball
carrier will accelerate.
 If the ball carrier has less momentum than the tackler,
he will be knocked backwards with a momentum equal to the
difference between the two players.
In many
instances, tacklers try to hold on to the ball carrier, and
the two may travel together. In these inelastic
collisions, the general reactions would be the same as
those above; however, in cases 2 and 3, the speeds at which
the combined players would move forward or backward would be
reduced. This reduction in speed is due to the fact that the
difference in momentum is now distributed over the combined
mass of the two players, instead of the mass of the one player
with the lesser momentum.
The Tackling
Process Coaches often tell their players to tackle a
runner low. In this way, the runner's feet will be rotated in
the air in the direction of the tackle. Let's look at this
closely:


Tackling a
runner low requires less force because the tackler is
farther away from the runner's center of
mass. 
The Language of
Physics
 Center of mass  The point in a body's
distribution of mass at which all of the mass can be
considered to be concentrated.
 Torque  A force that tends to produce
rotation or twisting
 Imagine that
the runner's mass is concentrated in a point called the
center of mass. In men, the center of mass is located
at or slightly above the navel; women tend to have their
center of mass below their navels, closer to their hips. All
bodies will rotate easiest about their center of mass. So, if
a force is applied on either side of the center of mass, the
object will rotate. This rotational force is called
torque, and is the product of the amount of force
applied and the distance from the center of mass at which the
force applied. Because torque is a product, the same torque
can be applied to an object at different distances from the
center of mass by changing the amount of force applied: Less
force is required farther out from the center of mass than
closer in. So, by tackling a runner low  far from the center
of mass  it takes less force to tackle him than if he were
tackled high. Furthermore, if a runner is hit exactly at his
center of mass, he will not rotate, but instead will be driven
in the direction of the tackle.
A lineman crouches low so that his
center of mass is closer to the ground. This makes it
hard for an opposing player to move him.

Similarly, coaches often advise linemen to stay low. This
brings their center of mass closer to the ground, so an
opposing player, no matter how low he goes, can only contact
them near their center of mass. This makes it difficult for an
opposing player to move them, as they will not rotate upon
contact. This technique is critical for a defensive lineman in
defending his own goal in the "red" zone, the last 10 yards
before the goal line.
We have only touched on some of the applications of physics
as they relate to football. Remember, this knowledge appears
to be instinctive; Most often, players and coaches don't
consciously translate the mechanics of physics into their
playing of the sport. But by making that translation, we can
understand and appreciate even more just how amazing some of
the physical feats on the football field really are. Also,
applying physics to football leads to better and safer
equipment, affects the rules of the sport, improves athletic
performance, and enhances our connection to the game.
For more information on football physics and related
topics, check out the links on the next page.
Lots More Information!
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