Two stones are projected simultaniously from a tower at different angles of projection with same speed $u$. The distance between two stones is increasing at constant rate $u$. Then the angle between the initial velocity vectors of the two stone is:

The correct option is C ${60}^{\circ }$
Let, $O$ is the posituon of the observer.

Let, the stones are at positions $\text{A}$ and $\text{B}$ at shown in the figure.


Given, velocity with which stones are prjected is,

$\frac{dx}{dt}=u.......\left(1\right)$

The distance between two stones is increasing at rate $u$,

$\frac{dl}{dt}=u.......\left(2\right)$

$\therefore x=l$ So,

$\angle BOA={60}^{\circ }$

Triangle $\text{BOA}$ would be an equilateral triangle and all angles inside it would be sixty degrees.

Hence, option (b) is correct answer.

Why this question? To make the student understand that in the reference frame in which relative acceleration is zero, the relative velocity remains constant and relative displacement increases at a constant rate.