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ANALYSIS OF A FOOTBALL PUNT

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ANALYSIS OF A FOOTBALL PUNT

David Bannard

TCM Conference

NCSSM 2005

- 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.

- 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?

- 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?

- 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

- 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.

- 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.

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

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

Initial Velocity V0

q

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

Initial Velocity V0

q

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

Initial Velocity V0

q

X(t)=V0t cos q

- 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

q

X(t) = V0t cos q

- 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

- 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

- 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°.

- 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?

- 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?

- 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 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.

Use a Spreadsheet and/or calculator to collect data.

Then analyze the data using data analysis techniques on a calculator

- Can we determine how the distance the ball will travel relates to the initial velocity and 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

- 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

- 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

- 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.

- 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

- 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

- 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 qand 0 = –16t2 + V0t sinq.
- Solve both equations for V0 and set them equal.

- 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 q, namely
- X = V0t cos qand 0 = –16t2 + V0t sinq.
- Solve both equations for V0 and set them equal.