Football Question: Help Solving Quadratic Equation

[HelpRequired]

*Note*: I have just started in my physics course and don't understand much of the material and I do require quite a bit of help but any given will be appriciated!

Please bare with my lack of understanding for this subject !

Homework Statement

Vy = 14m/s
Vx = 9m/s
a = -9.81m/s

Question is basically asking that, with the information given above, will the football kicked (starting at 0), clear a bar that's 3m tall and is 20m away.

Homework Equations

Teacher expects us to use the quadratic formula and use the formula (1/2a x Delta t^2 + V1 Delta t - Delta d) to replace Ax^2 + Bx + C.

The Attempt at a Solution

I tried to solve for Delta t first with this:

Delta t = 1/2 x (-9.81m/s) + 14m/s (Vy) - 3 (The Bar's height).

a = -4.905m/s
b = 14m/s
c = -3m/s

The two possible answers I got after using the quadratic formula was: 3.0544 and 1.964.

I don't know what to do from here or even if I did the first part right.

Please help !

 

[gneill]

Separate the horizontal and vertical components of the motion. The vertical component is influenced by gravity (so it has that (1/2)g^t term), while the horizontal motion has a constant velocity.

You've been given the initial Vx and Vy velocities. So if the horizontal (x) velocity is constant, can you find the time to reach the distance of the goalpost?

 

[AdrianMay]

That's right. Start by finding the time from the horizontal component. Then, figure out the vertical speed after that time: it's just going to reduce by the force of gravity times the time. Now you have the starting and final vertical velocities and you can average them to get the average vertical velocity. Multiply that by the time to get the height. So is it over or under the bar?

Trouble is, teacher wants them to do it the hard way. Do the above first so you know the answer you're expecting from the hard way.

Let's try to construct this parabola for height vs horizontal distance. You know that y=x^2 makes a parabola. It goes through (0,0), (1,1), (-1,1), (2,4), etc. Let's try and correct it until it fits the experiment. First problem: it's upside down. We want something with the bottom of the bowl pointing upwards, so change it to y=(-x)^2. Next problem: it looks like you're playing football underground so change it to y=(-x)^2+c where c is something we've yet to figure out. Next problem: the peak is at zero but that position has nothing to do with our problem - we don't know that the peak is going to be at the goal line, and it certainly isn't where we kick it. Let's accommodate that with another unknown called b which is the x coordinate of the peak in whatever coordinates we end up using: y=(-(x-b))^2+c. The only thing that could still be wrong is the width which is likely to depend on the force of gravity: y=(-(ax-b))^2+c.

Now let's think about these constants. c is the highest point it will reach. That's easy. You have the initial vertical velocity so just use kinetic and potential energies:

mgh = 1/2 m Vy^2
h = Vy^2/2g = 14^2 / (2*9.81) = ...

How about a? The second differential of our equation is 2a, and that must have something to do with the force of gravity. But the force of gravity is in terms of time, and our equation is in terms of distance, so we need a mapping from one to the other. That's just the horizontal velocity. If the horizontal velocity was 1m/s, the second diff should be g, but if it was 2m/s, the path would look straighter. So we want g/Vx for the second diff, which is 2a:

a = g/2Vx = 9.81/(2*9) = ...

Our equation now reads:

y=(-(ax-b))^2 + c
y=(-([g/2Vx]x-b))^2 + Vy^2/2g
(you can shove the actual numbers in now to make this look nicer)

Now we just want b, and we'll get it by making the graph fit the start condition where Vx=9m/s and Vy=14m/s. We'll pretend b is zero, find out the x where the slope is 14/9, then twiddle b until it's 20m.

Let's start by expanding and diffing the above:

y=(-(ax-b))^2 + c
=(b-ax)^2 + c
=b^2 - 2abx + a^2.x^2 + c

dy/dx = -2ab + 2a^2.x
= 2a(ax-b)
= 2.[g/2Vx].([g/2Vx].x - b)

14/9 = 2*[9.81/2*9].([9.81/2*9]*20 - b)
b = ...

Now you have numbers for a, b and c in a coordinate system where you kicked the ball at (0,0). That's not the same as the a,b and c in "y = ax^2 + bx + c". If you like, you can work out the teacher's a, b and c. Either way, you can just plug in the 20m and get y. Hopefully it's the same as you got by doing it the easy way.

Adrian.

 

[gneill]

Or you could just write

y(t) = Vy*t - (1/2)*g*t²

which the teacher gave him.

 

[AdrianMay]

But did he know why?