Kinematics- Velocity at Origin

The general equations for motion with constant acceleration are $\left\{\begin{array}{l} x = x_0 + v_{x,0}t + \tfrac12a_xt^2 \\ y = y_0 + v_{y,0}t + \tfrac12a_yt^2 \end{array}\right.$ In this case we have $x_0 = y_0 = 0$ , $a_x = 0$ , $a_y = -a$ , so that $\left\{\begin{array}{l} x = v_{x,0}t \\ y = v_{y,0}t - \tfrac12at^2 \end{array}\right.$ Write $t = x/v_{x,0}$ to eliminate $t$ in the second equation: $y = \frac{v_{y,0}}{v_{x,0}}x - \frac{a}{2v_{x,0}^2}x^2.$ Compare with the given form $y = px - qx^2$ to conclude $p = \frac{v_{y,0}}{v_{x,0}}, q = \frac{a}{2v_{x,0}^2}.$ Solve for the initial velocity components: $\left\{\begin{array}{l} v_{x,0} = \sqrt{\frac{a}{2q}}; \\ v_{y,0} = pv_{x,0}.\end{array}\right.$ The magnitude of the initial velocity is then $v_0 = \sqrt{v_{x,0}^2 + v_{y,0}^2} = \sqrt{(1+p^2)v_{x,0}^2} = \sqrt{(1+p^2)\frac{a}{2q}} = \sqrt{\frac{a(1+p2^2)}{2q}}.$