A batsman hits a ball with a velocity 'v', making an angle of $60^\circ$ with the vertical. After some time direction of velocity is making an angle of $60^\circ$ with the horizontal. The speed of the ball at this instant is
$\left[\cos(60^\circ) = \dfrac{1}{2}, \cos(30^\circ) = \dfrac{\sqrt{3}}{2}\right]$

The initial angle of the velocity with the vertical is $60^\circ$. Therefore, the initial angle with the horizontal is $90^\circ - 60^\circ = 30^\circ$.

The horizontal component of the initial velocity is $u_x = v \cos(30^\circ)$.

Let the speed of the ball be $v'$ when its direction of motion makes an angle of $60^\circ$ with the horizontal. The horizontal component of velocity at this instant is $v'_x = v' \cos(60^\circ)$.

In projectile motion, the horizontal component of velocity remains constant throughout the flight.

Equating the horizontal components of velocity:

$v' \cos(60^\circ) = v \cos(30^\circ)$

Substituting the given trigonometric values:

$v' \left(\dfrac{1}{2}\right) = v \left(\dfrac{\sqrt{3}}{2}\right)$

$v' = \sqrt{3}v$

Answer: $\sqrt{3}v$