A ball of mass 0.1 kg is thrown vertically upwards with an initial velocity of 80 m/s. Calculate the potential energy (i) half way up (ii) at its maximum height. What is its

At maximum height, the final velocity $v_f = 0 \text{ m/s}$. Using the kinematic equation $v_f^2 = v_0^2 + 2ad$:

$$0^2 = v_0^2 - 2gH_{max}$$ $$H_{max} = \frac{v_0^2}{2g}$$

Substitute the given values:

$$H_{max} = \frac{(80 \text{ m/s})^2}{2 \times 9.8 \text{ m/s}^2} = \frac{6400 \text{ m}^2/\text{s}^2}{19.6 \text{ m/s}^2}$$ $$H_{max} \approx 326.53 \text{ m}$$

Potential energy $PE = mgh$.

$$PE_{max} = m g H_{max}$$ $$PE_{max} = (0.1 \text{ kg})(9.8 \text{ m/s}^2)(326.53 \text{ m})$$ $$PE_{max} = 320 \text{ J}$$

Alternatively, by conservation of energy, the potential energy at maximum height is equal to the initial kinetic energy:

$$PE_{max} = \frac{1}{2} m v_0^2 = \frac{1}{2} (0.1 \text{ kg})(80 \text{ m/s})^2 = \frac{1}{2} (0.1)(6400) = 320 \text{ J}$$

The potential energy at its maximum height is $\boxed{320 \text{ J}}$.

The height half way up is $H_{half} = \frac{H_{max}}{2}$.

$$H_{half} = \frac{326.53 \text{ m}}{2} = 163.265 \text{ m}$$ $$PE_{half} = m g H_{half}$$ $$PE_{half} = (0.1 \text{ kg})(9.8 \text{ m/s}^2)(163.265 \text{ m})$$ $$PE_{half} = 160 \text{ J}$$

The potential energy half way up is $\boxed{160 \text{ J}}$.

As the ball leaves the ground, its velocity is its initial velocity $v_0$.
Kinetic energy $KE = \frac{1}{2} m v^2$.

$$KE_{ground} = \frac{1}{2} m v_0^2$$ $$KE_{ground} = \frac{1}{2} (0.1 \text{ kg})(80 \text{ m/s})^2$$ $$KE_{ground} = \frac{1}{2} (0.1)(6400)$$ $$KE_{ground} = 320 \text{ J}$$

The kinetic energy as it leaves the ground is $\boxed{320 \text{ J}}$.