A ball being shot from level ground with a twist

[mooha]

Not a literal twist! It's just a bit more complex.

Homework Statement

A ball is to be shot from level ground toward a wall at a distance "x". I am given the graph of the y component of velocity just as it would reach the wall, as a function of that distance x(m). Here's the equation of the line:y=-.5x+5, [0, 20]
What is the launch angle?

Homework Equations

v = v_o +at
x = x_o + V_ot + .5at²
v² = (v_o)² + 2a(x - x_o)

The Attempt at a Solution

So I wrote out all I knew about the x and y components
y: v_o = 5, a = -9.8, x_o = 0
x: a=0

I attempted to created a triangle of the horizantal and vertical components of velocity and the initial velocity as the hypotenuse solve with trig. I got
sin(θ)= (5)/(V)
tan(θ)= (5)/(v_ox)
No help.
I really am stuck! Using kinematics, you can't find the initial velocity because you don't know the time. You can't find the time, because you don't know where in the 20 m of the trajectory the ball hit the wall (according to my interpretation of the problem) Ah! If you could help I would be so grateful!
Thank you so much!

 

[voko]

x depends on time. Use this dependency.