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 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.