Starting at time t = 0 from the origin with speed 1m s ^ - 1 , a...

$y = \dfrac{{x}^{2}}{2}$

$\dfrac{dy}{dt} = \dfrac{1}{2} \times 2x\dfrac{dx}{dt}$

${v}_{y} = x{v}_{x}$

$\dfrac{d{v}_{y}}{dt} = \dfrac{dx}{dt} \times \dfrac{dx}{dt} + x\left(\dfrac{{d}^{2}{v}_{x}}{d{t}^{2}}\right)$

${a}_{y} = {v}_{x}^{2} + x{a}_{x}$

If $x = 0, {a}_{y} = {v}_{x}^{2} = 1,$ No matters what is the value of ${a}_{x}$ hence, Option $A$ is correct.

If ${a}_{x} = 0$ then $x = {v}_{x}t = t$

at $t = 1; x = 1m$

then angle made by velocity vector with $x$ -axis

$tan\theta = \dfrac{{v}_{y}}{{v}_{x}} = x = 1 ⇒ \theta = 45 °$