I'm currently pretty stumped on the following question.
Let $$\mathbf{r}(s) = \left(\frac{3}{5}\cos(s),\frac{4}{5}s,\frac{3}{5}\sin(s)\right)$$ define a 3D curve.
If a frictionless particle moves on this curve under gravity, and starts with speed $U$ from the point $\left(\frac{3}{5},0,0\right)$, what is the condition on $U$ such that the velocity is never $0$.
For my part, I have calculated the velocity $$\mathbf{v}(s) = \mathbf{r}'(s) = \left(-\frac{3}{5}\sin(s),\frac{4}{5},\frac{3}{5}\cos(s)\right) $$ (which results in an initial speed of $1$ - also confusing) and the acceleration $$\mathbf{a}(s)= \mathbf{v}'(s) = \left(-\frac{3}{5}\cos(s),0,-\frac{3}{5}\sin(s)\right),$$ without gravity but I am struggling to fit gravity into this picture. Should I put in a $-(0,0,g)$ into $\mathbf{a}(s)$? I've realised that this curve is parametrised in terms of arc-length, does this mean that I must do something else?
How would people suggest I interpret this question as I think that is where I am stuck.