I haven't done any kinematics in years, so I could be wrong.. (in which case I'll be happy to remove this answer!) but I don't think your approach will work. You've set the initial velocity well, but the distance as 52m. We know the vehicle has to travel 52 to meters from the ramp to get to the ground, but you haven't factored in any of the "above 52m" travelling time.
I'll let the initial velocity be denoted as $v_i$. I think what you need to do is calculate how high the vehicle goes up in the air, then calculate from that height, how long it takes to get back to the ground. Namely, let $t_1$ be the time it takes for the vehicle to stop travelling up, and start travelling down, and let $t_2$ be the time from this point to when the cart hits the ground. To compute $t_1$, we have \begin{align*} v_f^2 = v_i^2 + 2ad \end{align*} where $v_f = 0$, so we find \begin{align*} d = -v_i^2/(2a). \end{align*} Notice we are moving up, against gravity, so your value for $a$ should be -10, -9.8,-9.81, or whatever you guys use for gravity. Now that we know how high the cart travels, we can compute the time to get there. We have \begin{align*} v_f = v_i + at_1 \end{align*} and so \begin{align*} t_1 = -v_i/a. \end{align*} To compute $t_2$, we know we traveled $d$ meters above the ramp, so from here, our travel down will be $d + 52$ meters. Hence we have \begin{align*} d + 52 = v_it_2 + \frac{1}{2}at_2^2 \end{align*} (where as we are travelling with gravity, $a$ is now positive) and so \begin{align*} t_2 = \sqrt{\frac{2(d + 52)}{a}}. \end{align*} The cart travels in the air for a time of $t_1 + t_2$.