Velocity Augmentation Suit (Part 2)

From our previous problem:

$\large \displaystyle v_x = 10e^{(\alpha/150)t}$

$\large \displaystyle v_y = (1500/ \alpha) + (20 - 1500 / \alpha) e^{(\alpha /150)t}$

Integrating with respect to time, and considering that at $t=0$ both $x$ and $y$ are equal to $0$ :

$\color{#20A900} \boxed{\large \displaystyle x = (1500/ \alpha) e^{(\alpha/150)t}}$

$\color{#20A900} \boxed{\large \displaystyle y = (1500/ \alpha)t + (3000/ \alpha - 225000 / \alpha^2) e^{(\alpha /150)t}}$

We want $x$ to be $1000$ while $y$ is $0$ . So, for $x$ :

$\large \displaystyle (1500/ \alpha) e^{(\alpha/150)t} = 1000$

$\color{#20A900} (i) \ \boxed{\large \displaystyle e^{(\alpha/150)t} = 2\alpha /3}$

Also, for $y$ :

$\large \displaystyle (1500/ \alpha)t + (3000/ \alpha - 225000 / \alpha^2) e^{(\alpha /150)t} = 0$

Plugging (i) on this:

$\large \displaystyle (225000 / \alpha^2 - 3000/ \alpha) (2 \alpha /3) = (1500 / \alpha) t$

$\color{#20A900} (ii) \ \boxed{\large \displaystyle t = \frac{300 - 4\alpha}{3}}$

Plugging (ii) in (i):

$\large \displaystyle e^{\frac{(300-4\alpha) \alpha}{450} } = \frac{2 \alpha}{3}$

I've solved this numerically (if there's a different solution, I'm happy to see), and obtained:

$\color{#3D99F6} \boxed{\large \displaystyle \alpha = 68.74}$