Maximum Vertical Velocity of Pendulum Say there is a 10N pendulum held parallel to the ground. In a frictionless environment, it will continue swinging along a semicircular path. The pendulum tip should have the highest horizontal component of velocity at the very bottom $[ 90^o ]$, and lowest at the top $[ 0^o ]$ , $[ 180^o ]$. 
What can similarly be said about the vertical component of velocity? At what angle during its swing will the tip be traveling fastest downwards and fastest upwards?
At the top of the swing, all of the pendulum's velocity is in the vertical direction, but the total velocity is zero. At the bottom of the swing, there is no up/down component of motion, and vertical velocity is again zero. $[ 45^o ]$ and $[ 135^o ]$ seem reasonable, as they're halfway between. 
 A: Suppose the pendulum is held at a height of $H$ from the ground. Then by conservation of energy, at any time $t$:
$$\frac{1}{2}mv^2+mgh=mgH$$
Or:
$$v=\sqrt{2g(H-h)}$$
Using geometry, one can show that the vertical component of the velocity is given by $v\cos{\theta}$, hence:
$$v_{\textrm{ver}}=\cos{\theta}\sqrt{2g(H-h)}$$
More geometry reveals that $h=H(1-\sin{\theta})$, hence:
$$v_{\textrm{ver}}=\cos{\theta}\sqrt{2gH\sin{\theta}}$$
The derivative of this function is:
$$\frac{\textrm{d}v_{\textrm{ver}}}{\textrm{d}\theta}=\sqrt{2gH}\frac{\cos^2{\theta}-2\sin^2{\theta}}{2\sqrt{\sin{\theta}}}$$
Setting this equal to zero, we find that the nontrivial critical points, corresponding to the maxima, are:
$$\theta=\frac{1}{2}\arccos{\frac{1}{3}}$$
And:
$$\theta=180^{\circ}-\frac{1}{2}\arccos{\frac{1}{3}}$$
These correspond to approximately $35.3^{\circ}$ and $144.7^{\circ}$, respectively.
I have graphed the parent vertical velocity function and vertical lines at the maxima here:

Note that this graph is in radians.
