A ball is projected with a velocity, $10{\text{ms}}^{-1}$, at an angle of $60^{\circ }$ with the vertical direction. Its speed at the highest point of its trajectory will be:

Angle with vertical is $60^{\circ }$. Therefore, the angle with the horizontal is $\theta =90^{\circ }-60^{\circ }=30^{\circ }$.At the highest point of a projectile's trajectory, the vertical component of velocity is zero, and only the horizontal component remains.
$$v_h=u\cos \theta =10\cos 30^{\circ }=10\times \frac{\sqrt{3}}{2}=5\sqrt{3}{\text{ms}}^{-1}$$