Transcript ANALYSIS OF A FOOTBALL PUNT

ANALYSIS OF A FOOTBALL
PUNT
David Bannard
TCM Conference
NCSSM 2005
Opening thoughts
• Watching St. Louis, Atlanta playoff game,
the St. Louis punter punts a ball.
• At the top of the screen a hang-time of 5.1
sec. is recorded.
• In addition, I observed that the ball traveled
a distance of 62 yds.
What questions might occur
to us!
• How hard did he kick the ball?
• Asked another way, how fast was the ball
traveling when it left his foot?
• At what angle did he or should he have
kicked the ball to achieve maximum
distance?
• How much effect does the angle have on the
distance?
More Questions
• How much effect does the initial velocity
have on the distance?
• Which has more, the angle or the initial V?
• What effect does wind have on the punt?
Initial Analysis
• Most algebra students have seen the
equation h  16t 2 V t  H
0
0
• Suppose we assume the initial height is 0.
When the ball lands, h = 0, so we have
2
0

16t
 V0t or V0  16t
•
• In other words, a hang-time of 5.0 sec.
Would result from an initial velocity of 80
ft/sec
Is This Solution Correct?
• Note that this solution only considers
motion in one dimension, up and down.
• The graph of this equation is often
misunderstood, as students often think of
the graph as the path of the ball.
• To see the path the ball travels, the x-axis
must represent horizontal distance and the
y-axis vertical distance.
Two dimensional analysis
• Using vectors and parametric equations, we
can analyze the problem differently.
• We will let X(t) be the horizontal
component, I.e. the distance the ball travels
down the field, and Y(t) be the vertical
component, the height of the ball.
• Both components depend on the angle at
which the ball is kicked and the initial V.
Vector Analysis
• The horizontal component depends only on
V t and the cosine of the angle.
• The vertical component combines v t sinq
and the effects of gravity, –16t2.
0
0
Y(t)=–16t2+V0t sinq
q
X(t)=V0t cos q
Calculator analysis
• In parametric mode, enter the two
equations.
• X(t)=V0t cos q + Wt where W is Wind
• Y(t)=–16t2+V0t sin q + H0 where H0 is the
initial height.
• However we will assume W and H0 are 0
Initial Parametric Analysis
• Suppose that we start with t = 5 sec. and
V0=80 ft./sec.
• We need an angle, and most students
suggest 45° as a starting point.
• These values did not give the results that
were predicted by the original h equation.
• Try using a value of q=90°.
Trial and Error
• Assume that the kicking angle is 45°. Use
trial and error to determine the initial
velocity needed to kick a ball about 62
yards, or 186 feet.
• What is the hang-time?
New Questions
• 1) How is the distance affected by changing
the kicking angle?
• 2) How is the distance affected by changing
the initial velocity?
• 3) Which has more effect on distance?
Data Collection
• Collect two sets of data from the class
• Set 1: Hold the velocity constant at 80
ft/sec. And vary the angle from 30° to 60°.
• Set 2: Hold the angle constant at 45° and
vary the velocity from 60 ft/sec to 90 ft/sec.
Accuracy
• Accuracy will improve by making delta t
smaller. Dt = 0.05 is fast. Dt = 0.01 is more
accurate.
• Do we wish to interpolate?
• First estimate the hang-time with Dt = 0.1
• Use Calc Value to get close to the landing
place.
• Choose t and X at the last positive Y.
Using a Spreadsheet to collect
data.
Algebraic Analysis
• Can we determine how the distance the ball
will travel relates to the initial velocity anf
the angle. In particular, why is 45° best?
• X(t) = V0t cos q and Y(t) = –16t2 + V0t sin q
• When the ball lands, Y = 0, so
• –16t2 + V0t sin q  0 or t (–16t + V0 sinq) =
0
• So t = 0 or V0 sinq/16.
• But X(t) = V0t cos q
V sin q cosq
X(t) 
16
• Substituting gives
• Using the double angle
2 identity gives
2
0
V0 sin 2q
X(t) 
32
V0 2 sin 2q
X(t) 
32
• Finally, we have something that makes
sense.
• If V0 is constant, X varies as the sin of 2q,
which has a maximum at q = 45°.
• If q is constant, X varies as the square of V0.
Additional results
• How do hang-time and height vary with q
and V0?
• We already know the t = V0 sinq/16
• The maximum height occurs at t/2, so
 V sin q 
 V0 sin q 
Y max  16  0

V
sin
q


0
 32 
32 
2
2
2
2
2
16
1
V
sin
q
1
V
sin
q




2
2
0
0
Y max  V0 sin q  2   



 32
32 
32  2 
64
Final Question
If we know the hang-time, and distance, can
we determine V0 and q?
•
•
•
•
Given that when Y(t)=0, we know X(t) and t.
Therefore we have two equations in V0 and q, namely
X = V0t cos q
and 0 = –16t2 + V0t sinq.
Solve both equations for V0 and set them equal.
X
16t

t cos q sin q
sin q 16t 2
so

cosq
X
16t
and V0 
sin q
16t 2
or tan q 
X