The initial velocity of a projectile is given by u = (6 i + 8 j ) m...

In projectile motion, the horizontal component of velocity remains constant throughout the flight, while the vertical component changes due to gravity.

Given the initial velocity vector $\vec{u} = 6\hat{i} + 8\hat{j}$, the horizontal component is $u_x = 6 \text{ m s}^{-1}$ and the vertical component is $u_y = 8 \text{ m s}^{-1}$.

At the maximum height, the vertical component of velocity becomes zero ($v_y = 0$).

Therefore, the velocity at the maximum height is entirely horizontal: $\vec{v} = 6\hat{i} \text{ m s}^{-1}$.

The speed at the maximum height is the magnitude of this velocity, which is $6 \text{ m s}^{-1}$.

Answer: $6 \text{ m s}^{-1}$