Step 1: Find the Horizontal Velocity ($u_x$): At the highest point of a trajectory, the vertical velocity is zero. Only the horizontal component remains, which is constant throughout the flight ($v_{top} = u \cos \theta$).
Given Kinetic Energy at top ($KE_{top}$) = $200 \text{ J}$ and $m = 1 \text{ kg}$:
$$KE_{top} = \frac{1}{2} m (u \cos \theta)^2$$
$$200 = \frac{1}{2} (1) (u_x)^2 \implies u_x^2 = 400 \implies u_x = 20 \text{ m/s}$$
Step 2: Find the Initial Vertical Velocity ($u_y$): The maximum height ($H$) formula is $H = \frac{u_y^2}{2g}$. Taking $g = 10 \text{ m/s}^2$:
$$20 = \frac{u_y^2}{2(10)} \implies 20 = \frac{u_y^2}{20}$$
$$u_y^2 = 400 \implies u_y = 20 \text{ m/s}$$
Step 3: Calculate Resultant Initial Velocity ($u$): The total velocity from the ground is the vector sum of these components:
$$u = \sqrt{u_x^2 + u_y^2}$$
$$u = \sqrt{20^2 + 20^2} = \sqrt{400 + 400} = \sqrt{800}$$
$$u = 20\sqrt{2} \text{ m/s}$$