To solve this problem, we need to determine the speed of the ball at the highest point of its trajectory.$\\$Given:$\\$• Initial velocity $u = 10$ m/s$\\$• Angle with the vertical $= 60^\circ \\$First, we need to find the horizontal and vertical components of the initial velocity.$\\$The angle given is with the vertical, so the horizontal component $u_x$ is:$\\ u_x = u \cdot \sin(60^\circ) \\$The vertical component $u_y$ is:$\\ u_y = u \cdot \cos(60^\circ) \\$Substitute the known values:$\\ u_x = 10 \cdot \sin(60^\circ) \\ u_y = 10 \cdot \cos(60^\circ) \\$Using trigonometric values:$\\ \sin(60^\circ) = \frac{\sqrt{3}}{2} \\ \cos(60^\circ) = \frac{1}{2} \\$Calculate the components:$\\ u_x = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}$ m/s$\\ u_y = 10 \cdot \frac{1}{2} = 5$ m/s$\\$At the highest point of the trajectory, the vertical component of the velocity becomes zero,$\\$but the horizontal component remains unchanged.$\\$Therefore, the speed at the highest point is equal to the horizontal component:$\\$Speed at the highest point $= u_x = 5\sqrt{3}$ m/s$\\$Thus, the correct option is 5$\sqrt{3}$ m/s, which corresponds to Option 2.