# Equation of a badminton birdie

• Jan 2nd 2009, 04:14 PM
fkhnksaha
So the problem goes like this:

In an underhand serve, the birdie starts from a point 2m above the ground and travels a horizontal distance of 20m before it strikes the floor. The birdie travels in a parabolic path and reaches it's highest point after traveling a horizontal distance of 9m. Determine the equation of the parabola in the form $y-k=a(x-h)^2$.

So I know of course that $(x-h)$ becomes $(x-9)$. The vertex is $(9,k)$. $a$ would have to be negative, since the parabola opens downwards.

Any actual solving the equation has me stumped, Conics wasn't my best unit =\
• Jan 2nd 2009, 05:08 PM
Shyam
Quote:

Originally Posted by fkhnksaha
So the problem goes like this:

In an underhand serve, the birdie starts from a point 2m above the ground and travels a horizontal distance of 20m before it strikes the floor. The birdie travels in a parabolic path and reaches it's highest point after traveling a horizontal distance of 9m. Determine the equation of the parabola in the form $y-k=a(x-h)^2$.

So I know of course that $(x-h)$ becomes $(x-9)$. The vertex is $(9,k)$. $a$ would have to be negative, since the parabola opens downwards.

Any actual solving the equation has me stumped, Conics wasn't my best unit =\

Let the equation be

$y - k = a(x - h)^2$

It has maximum height at horizontal distance 9, so, h = 9 Put this in eqn,

$y - k = a(x - 9)^2$ ...............(1)

Now, since it starts off at initial height 2, so, point (0, 2) lies in eqn (1)

Also, birdie strikes the floor at horizontal distance 20, so, point (20, 0) lies on eqn (1) also.

Plug in these point in eqn (1) and solve to find the values of a and k and put these values in eqn (1), you will get the equation

$y - \frac{121}{20} = \frac{-1}{20}(x - 9)^2$

Please see attached graph for clarification.
Did you get it?
• Jan 2nd 2009, 05:27 PM
fkhnksaha
Quote:

Originally Posted by Shyam
Plug in these point in eqn (1) and solve to find the values of a and k and put these values in eqn (1), you will get the equation

$y - \frac{121}{20} = \frac{-1}{20}(x - 9)^2$

Sorry, I do not understand the process of solving for a and k with (0, 2) and (20, 0).

Are you talking about doing something like

$2-k=a(0-9)^2$ and $0-k=a(20-9)^2$ ?

If so, how would you continue from that?
• Jan 2nd 2009, 05:38 PM
Shyam
Quote:

Originally Posted by fkhnksaha
Sorry, I do not understand the process of solving for a and k with (0, 2) and (20, 0).

Are you talking about doing something like

$2-k=a(0-9)^2$ and $0-k=a(20-9)^2$ ?

If so, how would you continue from that?

You must know the basic algebra and calculation.

$0-k=a(20-9)^2$

$-k=a(11)^2$

$k = -121a$ .................................(2)

$2-k=a(0-9)^2$

$2-k=a(81)$ ........................(3)

now, put the value of k from eqn (2) to eqn(3) and solve for "a"

• Jan 2nd 2009, 06:01 PM
fkhnksaha
Quote:

Originally Posted by Shyam
You must know the basic algebra and calculation.

$0-k=a(20-9)^2$

$-k=a(11)^2$

$k = -121a$ .................................(2)

$2-k=a(0-9)^2$

$2-k=a(81)$ ........................(3)

now, put the value of k from eqn (2) to eqn(3) and solve for "a"