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ANALYSIS OF A FOOTBALL PUNT David Bannard TCM Conference NCSSM 2005

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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.

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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?

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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?

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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

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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.

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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.

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Vector Analysis The Ball leaves the foot with an initial velocity V 0 at an angle with the ground.

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Vector Analysis The Ball leaves the foot with an initial velocity V 0 at an angle with the ground. Initial Velocity V 0

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Vector Analysis The horizontal component depends only on V 0 t and the cosine of the angle. Initial Velocity V 0

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Vector Analysis The horizontal component depends only on V 0 t and the cosine of the angle. Initial Velocity V 0 X(t)=V 0 t cos

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Vector Analysis The horizontal component depends only on V 0 t and the cosine of the angle. The vertical component combines v 0 t sin and the effects of gravity, –16t 2. Initial Velocity V 0 X(t) = V 0 t cos

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Vector Analysis The horizontal component depends only on V 0 t and the cosine of the angle. The vertical component combines v 0 t sin and the effects of gravity, –16t 2. Initial Velocity V 0 X(t) = V 0 t cos Y(t) = –16t 2 + V 0 t sin

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Calculator analysis In parametric mode, enter the two equations. X(t)=V 0 t cos + Wt where W is Wind Y(t)=–16t 2 +V 0 t sin + H 0 where H 0 is the initial height. However we will assume W and H 0 are 0

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Initial Parametric Analysis Suppose that we start with t = 5 sec. and V 0 =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°.

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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?

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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?

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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.

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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.

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Use a Spreadsheet and/or calculator to collect data. Then analyze the data using data analysis techniques on a calculator

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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?

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X(t) = V 0 t cos and Y(t) = –16t 2 + V 0 t sin When the ball lands, Y = 0, so –16t 2 + V 0 t sin or t (–16t + V 0 sin ) = 0 So t = 0 or V 0 sin /16. But X(t) = V 0 t cos

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X(t) = V 0 t cos and Y(t) = –16t 2 + V 0 t sin When the ball lands, Y = 0, so –16t 2 + V 0 t sin or t (–16t + V 0 sin ) = 0 So t = 0 or V 0 sin /16. But X(t) = V 0 t cos Substituting gives

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X(t) = V 0 t cos and Y(t) = –16t 2 + V 0 t sin When the ball lands, Y = 0, so –16t 2 + V 0 t sin or t (–16t + V 0 sin ) = 0 So t = 0 or V 0 sin /16. But X(t) = V 0 t cos Substituting gives Using the double angle identity gives

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Finally, we have something that makes sense. If V 0 is constant, X varies as the sin of 2, which has a maximum at = 45°. If is constant, X varies as the square of V 0.

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Additional results How do hang-time and height vary with and V 0 ? We already know the t = V 0 sin /16 The maximum height occurs at t/2, so

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Additional results How do hang-time and height vary with and V 0 ? We already know the t = V 0 sin /16 The maximum height occurs at t/2, so

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Final Question If we know the hang-time, and distance, can we determine V 0 and ? Given that when Y(t)=0, we know X(t) and t. Therefore we have two equations in V 0 and, namely X = V 0 t cos and 0 = –16t 2 + V 0 t sin. Solve both equations for V 0 and set them equal.

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If we know the hang-time, and distance, can we determine V 0 and ? Given that when Y(t)=0, we know X(t) the distance and t, the hang-time. Therefore we have two equations in V 0 and, namely X = V 0 t cos and 0 = –16t 2 + V 0 t sin. Solve both equations for V 0 and set them equal.

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