A ball is projected from point A with velocity 20 ms^-1 at an angle 60° to the horizontal direction. At the highest point B of the path (as shown in figure), the velocity of the ball will be:

When the body is projected at an angle ‘$\theta$’ with the horizontal, its horizontal velocity will be constant.

At the highest point, vertical velocity is zero.So, it has only horizontal velocity, i.e., $\mathrm{Vcos\theta }.$

$20\times \cos {60}^{\circ }=20\times 1/2=10\mathrm{m}/\mathrm{s}\text{.}$