Why does dy = vit + ½at² give negative height for fountain water?

[dragon-kazooie]

Homework Statement

1. A fountain shoots water out of the ground at a 40° angle. It’s horizontal velocity is 2 m/s and the water lands 3 meters away. What is the maximum height that the water reaches during that time?

Homework Equations

D_y =v_it + ½ at²

and

V_fy² = V_iy² + 2ad_y

The Attempt at a Solution

First solve for the time in the air: t = d_x /v_x t = 3m/2m/s = 1.5s
Divide that in half to find the time at maximum height: t = 0.75s

Solve for initial vertical velocity:
Tan θ = V_y /V_x
Tan 40° = V_y /2m/s
V_y = 1.68 m/s

Now I get two different answers when I use different equations, and I don't see why. Using this one that I did in purple must be wrong, because it doesn't make sense for ½ at² to be a larger value than v_it and I get a negative number which can't be right. But I don't see where the mistake is! What am I doing wrong?

d_y =v_it + ½ at²
d_y =1.68 × 0.75s + ½ -9.8m/s² × (0.75s)²
d_y = 1.26 - 2.75
d_y = -1.49m

If I use this formula in blue I get a different final answer which seems more correct:
V_fy² = V_iy² + 2ad_y
0 = 1.678² + 2 × -9.8 × d_y
0 = 2.816 + 2 × -9.8 × d_y
-2.816 = -19.6 d_y
0.143 = d_y

Is the one in blue correct? What did I do wrong with the one in purple?

Thank you in advance!

 

[haruspex]

@dragon-kazooie , your first method is correct. You got a crazy answer because the given conditions are impossible. There is no way that a jet of only 2m/s can reach 3m horizontally, even at 45 degrees.
Your second method appeared to work because you do not need the 3m range information, and you effectively ignored it.

 

[dragon-kazooie]

Thank you @haruspex! That was driving me crazy!