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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.

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ANALYSIS OF A FOOTBALL PUNT

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

  2. 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.

  3. 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?

  4. 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?

  5. 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

  6. 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.

  7. 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.

  8. Vector Analysis • The horizontal component depends only on V0t and the cosine of the angle. • The vertical component combines v0t sinq and the effects of gravity, –16t2. Initial Velocity V0 Y(t)=–16t2+V0t sinq q X(t)=V0t cos q

  9. 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

  10. 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°.

  11. 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?

  12. 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?

  13. 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.

  14. 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.

  15. Using a Spreadsheet to collect data.

  16. 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?

  17. 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 • Substituting gives • Using the double angle identity gives

  18. 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.

  19. 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

  20. Final QuestionIf 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.

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