A small ball is projected with initial speed $u$ and at an angle $\theta$ with horizontal from ground. The de-Broglie wavelength of ball at the moment its velocity vector becomes perpendicular to initial velocity vector is

If the velocity vector of the projectile at time $t$ is perpendicular to its initial velocity vector, the dot product of the velocity vectors at these two instants is zero.

$$\begin{gathered} \Rightarrow \left[u\cos \theta \hat{\mathrm{i}}+u\sin \theta \hat{\mathrm{j}}\right]·\left[u\cos \theta \hat{\mathrm{i}}+\left(u\sin \theta -gt\right)\hat{\mathrm{j}}\right]=0 \\ \Rightarrow u^2{\cos }^2\theta +u\sin \theta (u\sin \theta -gt)=0 \\ \Rightarrow t=\frac{u}{g\sin \theta } \end{gathered}$$

The vector is perpendicular to initial velocity vector at  $t=\frac{u}{g\sin \theta }$ and at this instant its speed is

$v=\sqrt{{\left(u\cos \theta \right)}^2+{\left(u\sin \theta -g\frac{u}{g\sin \theta }\right)}^2}=u\cot \theta$

Since, $\lambda =\frac{h}{mv}$,

$\Rightarrow \lambda =\frac{h}{mu\cot \theta }=\left(\frac{h}{mu}\right)\tan \theta$