# Thread: Modeling an NFL punter's kick with parametric equations?

1. ## Modeling an NFL punter's kick with parametric equations?

Hello,

My pre-cal teacher has been working on parametric equations with us recently and I am quite confused on the subject.

I understand that parametric equations, unlike standard functions, apply a parameter, most often t for time. And, one gives t as input, and receives output from x and y. Not sure if this is always the case, but it has been what I've seen most often.

I checked out the wikipedia article on parametric equations and the kinds of problems I usually have problems with are the ones explained at the top, "A simple kinematical example is when one uses a time parameter to determine the position, velocity, and other information about a body in motion."

OK, so I have the following problem:

An NFL punter at the 15-yard line kicks a football with an initial velocity of 95 feet per second at an angle of elevation of 18°. Let t be the elapsed time since the football is kicked.

a) Write the parametric equations of this problem.

b) With t = 0 at the time of the kick, how many seconds is the ball in the air?

c) Where does the ball land given the kick is straight and from the 15 yard line?

So, my biggest problem is finding the parametric equations described in part a. Once I have that figured out, I should be able to do parts b and c.

Now, should I create a non-parametric model first, and then parametricize that model?

Also, do I need to know some information other than what is given?

We are given:

Starting Position: 45-foot line
Initial Velocity: 95 feet/second (isn't there a formula describing this as v not?)
Angle of Elevation: 18°

Any help is appreciated!

Thanks!

2. The vertical component of the velocity, which is affected by gravity is 95sin18 (degrees) = 29.36 ft/sec. The horzontal component, 95 cos 18 = 90.35 ft/sec is not affected by gravity. The parameter is, as you suggest, the time (t). The parametric equations are therefore the equations of motion in the vertical and horizontal directions. The vertical equation of motion is given by y (height) = 29.36t - 16t^2 (t-squared). The horizontal motion is given by x = 90.35t. Solving the first equation for t = 0 gives you two solutions, t = 0 (the instant the ball is kicked) and t = 1.835 seconds (when the ball hits the ground). The horizontal equation tells you that the ball will travel 1.835 X 90.35 = 165.8 Ft.

3. I think starting with a parametric model is best here.
Firstly, you should decompose (not sure if that's the right word) the initial velocity in two vectors, one along the x-axis, one along the y-axis. These are the speeds in the respective direction.
You can do this by drawing a triangle with an angle of 18 degrees and an hypothenuse of length 45 and then calculate the two sides.

then, $y = speed_{y}*t - 0.5*g*t^2$
and $x = speed_{x}*t$
I got the formula's by altering some of the most basic physics formula's. If you don't get how I did that, please say so and I'll explain how I got there.

4. Originally Posted by Pim
I think starting with a parametric model is best here.
Firstly, you should decompose (not sure if that's the right word) the initial velocity in two vectors, one along the x-axis, one along the y-axis. These are the speeds in the respective direction.
You can do this by drawing a triangle with an angle of 18 degrees and an hypothenuse of length 45 and then calculate the two sides.

then, $y = speed_{y}*t - 0.5*g*t^2$
and $x = speed_{x}*t$
I got the formula's by altering some of the most basic physics formula's. If you don't get how I did that, please say so and I'll explain how I got there.
So, when you say decompose, do you mean the opposite of placing a vector in component form? I think I get the purpose, though. Why a hypotenuse of 45, though?

Let me see if I can get that down first before I digest the rest.

5. Indeed, that's what I meant by decomposing. The 45 is a misread on my part. (I meant the velocity of 95 feet/second)

6. ## football punt

Hello qcom,

Here is a simpler solution,
Using professorcools upward velocity of 29.36 FPS and the fact that the upward velocity will declerate to zero and then fall back to the starting elevation

V inital - gt =0
29.36 - 32,2 t =0
t= 29.36/32.2=.912 sec
ball in air 2x .912 =1.824 sec
Horizontal distance = 90.35 x 1.824 = 164.8 ft

bjh