Show that the potential energy gained by a body of mass m lifted vertically through a height h is E = mgh. Calculate the height to which a 350 kg ball must be raised to deliver 10 kJ of energy.

To show that the potential energy gained by a body of mass $m$ lifted vertically through a height $h$ is $E = mgh$:
When a body is lifted, work is done against the force of gravity.
The force of gravity (weight) acting on the body is given by:

$$F = mg$$

Where $m$ is the mass of the body and $g$ is the gravitational field strength.
The work done ($W$) in lifting the body through a vertical height $h$ is the product of the force and the distance moved in the direction of the force:

$$W = F \times h$$

Substituting the force $F = mg$:

$$W = mg \times h$$

This work done against gravity is stored in the body as gravitational potential energy ($E$).
Therefore, the potential energy gained by the body is:

$$E = mgh$$

This completes the derivation.

Given:
Mass of the ball $m = 350 \text{kg}$
Energy to be delivered $E = 10 \text{kJ} = 10 \times 1000 \text{J} = 10000 \text{J}$
Gravitational field strength $g = 9.8 \text{m s}^{-2}$

The energy delivered to the wall comes from the gravitational potential energy of the ball.

$$E = mgh$$

To find the height $h$:

$$h = \frac{E}{mg}$$ $$h = \frac{10000 \text{J}}{350 \text{kg} \times 9.8 \text{m s}^{-2}}$$ $$h = \frac{10000}{3430} \text{m}$$ $$h \approx 2.915 \text{m}$$

The ball must be raised to a height of $\boxed{\text{2.92 m}}$.

Assuming the ball does not lose any energy to the surroundings, the potential energy at its maximum height is converted entirely into kinetic energy at the moment of impact.
The kinetic energy ($KE$) at impact is equal to the energy delivered:

$$KE = E = 10000 \text{J}$$

The formula for kinetic energy is:

$$KE = \frac{1}{2}mv^2$$

Where $v$ is the velocity at impact.

$$10000 \text{J} = \frac{1}{2} \times 350 \text{kg} \times v^2$$ $$10000 = 175 v^2$$ $$v^2 = \frac{10000}{175}$$ $$v^2 \approx 57.1428$$ $$v = \sqrt{57.1428}$$ $$v \approx 7.559 \text{m s}^{-1}$$

The velocity at the moment of impact is $\boxed{\text{7.56 m s}^{-1}}$.