A ball is thrown vertically upwards with an initial velocity of 20 m/s. Neglecting air resistance, calculate the time it takes to return to the thrower's hand. (Take g = 10 m/s2)

Using the kinematic equation $v = u + at$:
$$
0 = 20 + (-10)t_{up}
$$
$$
10t_{up} = 20
$$
$$
t_{up} = \frac{20}{10}
$$
$$
t_{up} = 2 \text{ s}
$$

The time to go up is equal to the time to come down.  
Total time, $T = 2 \times t_{up}$
$$
T = 2 \times 2 \text{ s}
$$
$$
T = 4 \text{ s}
$$
The correct option is **B**.  
The final answer is $\boxed{\text{4 s}}$.

$$
\eta = \frac{\text{MA}}{\text{VR}}
$$

$$
\text{MA} = \eta \times \text{VR}
$$

$$
\text{MA} = 0.8 \times 5
$$
$$
\text{MA} = 4.0
$$
The correct option is **C**.  
The final answer is $\boxed{4.0}$.

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

$$
\text{KE} = \frac{1}{2}(10 \text{ kg})(5 \text{ m/s})^2
$$

$$
\text{KE} = \frac{1}{2}(10 \text{ kg})(25 \text{ m}^2/\text{s}^2)
$$
$$
\text{KE} = 5 \times 25 \text{ J}
$$
$$
\text{KE} = 125 \text{ J}
$$
The correct option is **C**.  
The final answer is $\boxed{125 \text{ J}}$.