In horizontal level, ground to ground projectile if at any instant velocity becomes perpendicular to initial velocity, then what can you say about projection angle with horizontal?

Velocity at any time, $\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{g}t$

$\Rightarrow \overrightarrow{v}=u\cos \theta \hat{i}+(u\sin \theta -gt)\hat{j}$

Let at any time, this velocity becomes perpendicular to initial velocity. Then $\overrightarrow{v}\cdot \overrightarrow{u}=0$

Solve to get $t=\frac{u}{g\sin \theta }$

Now t should be less than/equal to time of flight (T).

$\text{So} t\le T$

$\begin{array}{l}\frac{u}{g\sin \theta }\le \frac{2u\sin \theta }{g}\Rightarrow {\sin }^2\theta \ge \frac{1}{2}\\ \Rightarrow \sin \theta \ge \frac{1}{\sqrt{2}}\Rightarrow \theta \ge {45}^{\circ }\end{array}$