A worker accidentally drops a hammer of a mass of 1 kg from point X, which is at the edge of a flat roof of a building that is 35 m tall. Y is a point on the ground where the hammer lands, as shown in the diagram below. Ignore the effects of air resistance.

The mass of the hammer is $m = 1 \text{ kg}$.
The height of point X is $h = 35 \text{ m}$.
The acceleration due to gravity is $g = 9.8 \text{ m/s}^2$.
The formula for gravitational potential energy is $E_p = mgh$.

The gravitational potential energy of the hammer at point X is $\boxed{\text{343 J}}$.

5.3: The principle of conservation of mechanical energy states that in an isolated system where only conservative forces (like gravity) do work, the total mechanical energy (the sum of kinetic and potential energy) remains constant.

Since air resistance is ignored, mechanical energy is conserved.
Total mechanical energy at X = Total mechanical energy at Y

At point X, the hammer is dropped, so its initial velocity is $v_X = 0 \text{ m/s}$.
The height is $h_X = 35 \text{ m}$.

So, $E_{mech, X} = 343 \text{ J} + 0 \text{ J} = 343 \text{ J}$.

At point Y, the hammer is on the ground, so its height is $h_Y = 0 \text{ m}$.
Let the speed at point Y be $v_Y$.

The speed of the hammer at the instant it strikes the ground at point Y is $\boxed{\text{26.19 m/s}}$.