A particle is projected from the ground with a speed of 40 m s^ -1...

Let the point of projection be the origin. The initial velocity vector $\vec{u}$ can be resolved into horizontal and vertical components:

$\vec{u} = u \cos \theta \hat{i} + u \sin \theta \hat{j}$

$\vec{u} = 40 \cos 30^{\circ} \hat{i} + 40 \sin 30^{\circ} \hat{j} = 20\sqrt{3} \hat{i} + 20 \hat{j} \text{ m s}^{-1}$

At the highest point of the trajectory, the vertical component of velocity becomes zero, while the horizontal component remains constant.

Velocity at the highest point, $\vec{v} = 20\sqrt{3} \hat{i} \text{ m s}^{-1}$

The change in velocity vector is $\Delta \vec{v} = \vec{v} - \vec{u}$

$\Delta \vec{v} = (20\sqrt{3} \hat{i}) - (20\sqrt{3} \hat{i} + 20 \hat{j}) = -20 \hat{j} \text{ m s}^{-1}$

The magnitude of the change in velocity is $|\Delta \vec{v}| = 20 \text{ m s}^{-1}$.

Answer: $20 \text{ m s}^{-1}$