A ball is projected with a velocity, 10 ms ^ - 1 , at an angle of...

The speed of a projectile at highest point is equal to the component of velocity of projection along the horizontal direction. The angle given with the vertical is $60 °$ therefore, the angle with the horizontal will be $30 °$.

Therefore,

$V = ucos\theta$ $⇒ V = 10\cos 30 ° = 10 \times \dfrac{\sqrt{3}}{2}$

$⇒ V = 5\sqrt{3}m{s}^{- 1}$