20. A ball of mass $0.1 \text{ kg}$ is thrown vertically upwards with an initial velocity of $80 \text{ m/s}$. Calculate the potential energy (i) half way up (ii) at its maximum height. What is its kinetic energy as it leaves the ground? Given: Mass $m = 0.1 \text{ kg}$ Initial velocity $v_0 = 80 \text{ m/s}$ Acceleration due to gravity $g = 9.8 \text{ m/s}^2$ (assumed) Step 1: Calculate the maximum height ($H_{max}$). 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: $$ H_{max} = \frac{(80 \text{ m/s})^2}{2 \times 9.8 \text{ m/s}^2} = \frac{6400 \text{ m}^2/\text{s}^2}{19.6 \text{ m/s}^2} $$ $$ H_{max} \approx 326.53 \text{ m} $$ Step 2: Calculate the potential energy at its maximum height. 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 $\boxed{320 \text{ J}}$. Step 3: Calculate the potential energy half way up. 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 $\boxed{160 \text{ J}}$. Step 4: Calculate its kinetic energy as it leaves the ground. 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 $\boxed{320 \text{ J}}$.