Related to the information contained in $a = v \frac {dv}{ds}$ While studying kinematics I came to the definition of acceleration which is $a = \frac {dv}{dt}$. But from this equation we can derive that $ a = v \frac {dv}{ds} $ which when I evaluate at $v=0ms^{-1}$ (here as I know $v$ is the instantaneous velocity) gives $a=0$ which doesn't seem to be true, for example in the case of projectile at maximum height. So I think that what I'm doing is wrong somewhere but I can't figure out what it is. Can you please help tell what I doing wrong here? Are there any assumptions taken while deriving the given equation for which I'm not accounting for over here? 
 A: If you examine the equation $ a = v \frac{dv}{ds} $ when $v \rightarrow 0$, then it turns out that, $ \frac {dv}{ds} \rightarrow -\infty$. So when $v $ is exactly $0$, the acceleration takes the indeterminate form of $a=(0)(-\infty)$. Let me explain why.
As the velocity tends to $0$, the small change in the displacement i.e. $ds$ also tends to zero (it gets closer to zero than it was ever before). This can be seen by using the equation, $$v=\frac{ds}{dt} \Rightarrow vdt=ds$$
So clearly when $v=0 \Rightarrow ds=0$.
Caution:- I am not saying that $ds$ tends to $0$ (well, it always tends to zero, at least for our scenario), but here when $v=0$, it has an exact  value of $0$.
Thus now $\frac{dv}{ds}$ is no longer a $\frac{infinitesimal}{infinitesimal}$ form, but it is a $\frac{infinitesimal}{0}$ form which is equal to $±\infty$ (the sign of the infinity depends upon the sign of the infinitesimal quantity).
And as explained in the comments, the values of $\frac{dv}{ds}$ approaches $-\infty$ as the projectile reaches the peak and when the projectile is on its way down, the value of $\frac{dv}{ds}$ goes from $+\infty$(at the peak) to some finite value.
P.S. Many things might seem too "disgusting" to people who are passionate about mathematics as I have written this answer not using the appropriate mathematical rigour. I have used some vague, yet intuitive vocabulary in this answer. And that's intentional! It is because I want you to get the feel for it rather than getting stuck between evaluating limits.
A: If you have a parabolic trajectory
$$
s = \frac{1}{2} g t^2
$$
with velocity
$$
v = \frac{ds}{dt} = gt
$$
then note
$$
v = \sqrt{2gs}
$$
and
$$
\frac{dv}{ds} = \frac{\sqrt{g}}{\sqrt{2s}}.
$$
Note that the function $1/\sqrt{s}$ is infinite at $s = 0$. So the equation
$$
a = v \frac{dv}{ds}
$$
is true at $t = 0$, A.K.A. $s = 0$, when $v = 0$ because $\frac{dv}{ds}$ blows up in just the right way:
$$
a = v \times \frac{dv}{ds} =  \sqrt{2gs} \times \frac{\sqrt{g}}{\sqrt{2s}} = g.
$$
