During a projectile motion, the velocity

In projectile motion, the acceleration is constant and acts vertically downwards, $\vec{a} = -g \hat{j}$.

The velocity of the projectile at any instant $t$ is given by $\vec{v} = u \cos \theta \hat{i} + (u \sin \theta - gt) \hat{j}$.

For velocity to be perpendicular to acceleration, their dot product must be zero.

$\vec{v} \cdot \vec{a} = 0$

$-g(u \sin \theta - gt) = 0$

$t = \dfrac{u \sin \theta}{g}$

This time corresponds to the highest point of the trajectory. Since there is only one such instant during the motion, the velocity is perpendicular to the acceleration for a single instant only.