Starting from scratch I get the following.
let $c=-0.0015$
let $\dot{u} = \dfrac{du}{dt}$, i.e. a dot above a variable represents it's time derivative.
Note that $x$ is moving in the negative direction.
$y = c x^2 + 15$
$\dfrac{dy}{dx} = 2cx$
$\dot{y} = \dfrac{dy}{dx}\dot{x} = 2 c x \dot{x}$
$75^2 = \dot{x}^2 + \dot{y}^2 = (1+4 c^2 x^2)\dot{x}^2$
$v_x = \dot{x} = -\dfrac{75}{\sqrt{1+4c^2 x^2}}$
$v_y = \dot{y} = \dfrac{150 c x}{\sqrt{1+4 c^2 x^2}}$
$a_x = \dot{v_x} = \dfrac{dv_x}{dx}v_x$
$a_y = \dot{v_y} = \dfrac{d v_y}{dx} v_x$
$a_x = -\dfrac{600 c^2 x}{\left(4 c^2 x^2+1\right)^2} \dfrac{75}{\sqrt{1+4c^2 x^2}}=-\dfrac{45000 c^2 x}{\left(4 c^2 x^2+1\right)^3}$
$a_y = -\dfrac{150 c}{\left(4 c^2 x^2+1\right)^{3/2}}\dfrac{75}{\sqrt{1+4c^2 x^2}}= -\dfrac{11250 c}{\left(4 c^2 x^2+1\right)^{5/2}}$
plugging in $c$ and $x=50$ we obtain
$\vec{a} = (a_x,a_y) = \{-4.7356,-15.9619\}~ft/s^2$
$\vec{v} = \left<|v|\cos{\theta},|v|\sin{\theta}\right>$, where $\theta$ is the angle relative to the positive x-axis of the line tangent to the path at $x=50$ ...
$y' = -0.003x \implies y'(50) = -0.15 \implies \theta = \pi + \arctan(-0.15)$
$\vec{v} \approx \left<-74.17,11.13\right>$
$\vec{a} = \left<\dfrac{v^2}{r}\cos{\phi},\dfrac{v^2}{r}\sin{ \phi}\right>$, where $r = \dfrac{\left[1+(y')^2\right]^{3/2}}{|y''|}$ and $\phi = \theta + \dfrac{\pi}{2}$
$r = \dfrac{(1+0.0225)^{3/2}}{.003} \approx 344.64 \, ft$
$\vec{a} = \dfrac{75^2}{r} \approx 16.32 \, ft/s^2$
$a_x = \dfrac{75^2}{r} \cos{\phi} \approx -2.42 \, ft/s^2$
$a_y = \dfrac{75^2}{r} \sin{\phi} \approx -16.14 \, ft/s^2$