Find Initial Velocity from Angle and Horizontal Distance

[IcyDuck]

Homework Statement

"An athlete executing a long jump leaves the ground at a 30.0° angle and travels a horizontal distance 8.00m. What was the take-off speed?"

Let
$x=$ horizontal distance
$v=$ velocity
$a=$ acceleration

Known:
$x=8 m$
$x_0=0 m$
$a_y=-9.8 m/s^2$
$v_{0x}=v_0cos(30°)$
$v_{0y}=v_0sin(30°)$

Solve for $v_0$

Homework Equations

The only acceleration present is gravitational acceleration, so constant acceleration formulas are valid here. I decided to work with the horizontal component, because I was given the horizontal distance.
$v_x^2=v_{0x}^2+2a_x(x-x_0)$
$x=x_0+v_{0x}t+1/2a_xt^2$
$x=x_0+1/2(v_{0x}+v_x)t$

The Attempt at a Solution

I decided that since the athlete lands, and thus stops moving, at $x=8,$ the $x$ component of the final velocity would be $v_x=0.$ Given this and the values given above, I began plugging my values into the first relevant equation. But that led to $v_0=0$!

$v_x^2=v_{0x}^2+2a_x(x-x_0)$

$v_0x=v_0cos(30°).$

$0=(v_0cos(30°))^2,$

$v_0=0$


On my next attempt I went out on a bit of a limb and used the third relevant equation to derive time $t,$ which I knew would contain the unsolved variable $v_0.$ I got this and plugged this into the second relevant equation, which made the two $v_{0x}$ variables cancel out.

$x=x_0+v_0t+1/2a_xt^2$

$t=16/v_0$

$x=x_0+v_{0x}t+1/2a_xt^2$

$8=0+v_{0x}(16/v_{0x}),$


There's something I'm not taking into consideration (and I might be making this harder than it needs to be), but I don't know what. Any hints as to where I should be directing my thinking on this?

 

[rude man]

You are not taking the y displacement under consideration. And there is no x-acceleration. The only accelerative agent is gravity.

So write the equations for the x and y coordinates.