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ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005
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ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Mar 30, 2015

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Page 1: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

ANALYSIS OF A FOOTBALL PUNT

David Bannard

TCM Conference

NCSSM 2005

Page 2: 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.

Page 3: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

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?

Page 4: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

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?

Page 5: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Initial Analysis

• Most algebra students have seen the equation

• Suppose we assume the initial height is 0. When the ball lands, h = 0, so we have• • In other words, a hang-time of 5.0 sec.

Would result from an initial velocity of 80 ft/sec

h=−16t2 +V0t+H0

0 =−16t2 +V0t     or   V0 =16t

Page 6: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

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.

Page 7: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

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.

Page 8: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Vector Analysis

• The Ball leaves the foot with an initial velocity V0 at an angle with the ground.

Page 9: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Vector Analysis

• The Ball leaves the foot with an initial velocity V0 at an angle with the ground.

Initial V

elocity V0

Page 10: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Vector Analysis

• The horizontal component depends only on V0t and the cosine of the angle.

Initial V

elocity V0

Page 11: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Vector Analysis

• The horizontal component depends only on V0t and the cosine of the angle.

Initial V

elocity V0

X(t)=V0t cos

Page 12: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Vector Analysis

• The horizontal component depends only on V0t and the cosine of the angle.

• The vertical component combines v0t sin and the effects of gravity, –16t2.

Initial V

elocity V0

X(t) = V0t cos

Page 13: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Vector Analysis

• The horizontal component depends only on V0t and the cosine of the angle.

• The vertical component combines v0t sin and the effects of gravity, –16t2.

Initial V

elocity V0

X(t) = V0t cos

Y(t) = –16t2 + V0t sin

Page 14: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Calculator analysis

• In parametric mode, enter the two equations.

• X(t)=V0t cos + Wt where W is Wind

• Y(t)=–16t2+V0t sin + H0 where H0 is the initial height.

• However we will assume W and H0 are 0

Page 15: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

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 =90°.

Page 16: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

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?

Page 17: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

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?

Page 18: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

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.

Page 19: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Accuracy• Accuracy will improve by making delta t

smaller. t = 0.05 is fast. t = 0.01 is more accurate.

• Do we wish to interpolate?• First estimate the hang-time with t = 0.1• Use Calc Value to get close to the landing

place.• Choose t and X at the last positive Y.

Page 20: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Use a Spreadsheet and/or calculator to collect data.

Then analyze the data using data analysis techniques on a calculator

Page 21: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Algebraic Analysis

• Can we determine how the distance the ball will travel relates to the initial velocity and the angle. In particular, why is 45° best?

Page 22: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

• X(t) = V0t cos and Y(t) = –16t2 + V0t sin • When the ball lands, Y = 0, so

• –16t2 + V0t sin   0  or t (–16t + V0 sin) = 0

• So t = 0 or V0 sin/16.

• But X(t) = V0t cos

Page 23: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

• X(t) = V0t cos and Y(t) = –16t2 + V0t sin • When the ball lands, Y = 0, so

• –16t2 + V0t sin   0  or t (–16t + V0 sin) = 0

• So t = 0 or V0 sin/16.

• But X(t) = V0t cos

• Substituting gives X(t) =

V02 sin cos

16

Page 24: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

• X(t) = V0t cos and Y(t) = –16t2 + V0t sin • When the ball lands, Y = 0, so

• –16t2 + V0t sin   0  or t (–16t + V0 sin) = 0

• So t = 0 or V0 sin/16.

• But X(t) = V0t cos

• Substituting gives

• Using the double angle identity gives

X(t) =V0

2 sin cos16

X(t) =V0

2 sin232

Page 25: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

• Finally, we have something that makes sense.

• If V0 is constant, X varies as the sin of 2, which has a maximum at = 45°.

• If is constant, X varies as the square of V0.

X(t) =V0

2 sin232

Page 26: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Additional results

• How do hang-time and height vary with and V0?

• We already know the t = V0 sin/16

• The maximum height occurs at t/2, so

Page 27: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Additional results

• How do hang-time and height vary with and V0?

• We already know the t = V0 sin/16

• The maximum height occurs at t/2, so

Y max =−16V0 sin

32⎛⎝⎜

⎞⎠⎟

2

+V0 sinV0 sin

32⎛⎝⎜

⎞⎠⎟

Ymax=V02 sin2

−16322 +

132

⎛⎝⎜

⎞⎠⎟=V0

2 sin232

12

⎛⎝⎜

⎞⎠⎟=V0

2 sin264

Page 28: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

Final QuestionIf we know the hang-time, and distance, can

we determine V0 and ?• Given that when Y(t)=0, we know X(t) and t.

• Therefore we have two equations in V0 and , namely

• X = V0t cos and 0 = –16t2 + V0t sin.

• Solve both equations for V0 and set them equal.

Page 29: ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005.

If we know the hang-time, and distance, can we determine V0 and ?

• Given that when Y(t)=0, we know X(t) the distance and t, the hang-time.

• Therefore we have two equations in V0 and , namely

• X = V0t cos and 0 = –16t2 + V0t sin.

• Solve both equations for V0 and set them equal.X

t cos=

16tsin

so sincos

=16t2

X    or   tan =

16t2

X

and V0 =16tsin