Answer
$y=\frac{b}{2V}x^{3}$
Work Step by Step
Please see the attached image first.
We can apply the equation $S=ut+\frac{1}{2}at^{2}$ and apply this to both x, y (+) directions as follows.
$\rightarrow S=ut+\frac{1}{2}at^{2}$
Let's plug known values into this equation.
$x=Vt+0$
$x=Vt$
$t=\frac{x}{V}-(1)$
$\uparrow S=ut+\frac{1}{2}at^{2}$
Let's plug known values into this equation.
$y=0+\frac{1}{2}a(t)\space t^{2}$
$y=\frac{bt^{3}}{2}-(2)$
$(1)=\gt(2)$
$y=\frac{b}{2}(\frac{x}{V})^{3}$
$y=\frac{b}{2V^{3}}\space x^{3}\space -trajectory\space equation$
