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.