$(a)$ State Newton's second law of motion and show that the first law of motion can be mathematically derived from the mathematical expression for the second law of motion.
$(b)$ A stone dropped from a window reaches the ground in $0.5 \, s$ (given $g = 10 \, m \, s^{-2}$):
$(i)$ Calculate the speed just before it hits the ground.
$(ii)$ What is the average speed during $0.5 \, s$?
$(iii)$ Calculate the height of the window from the ground.

(N/A) The second law of motion states that, "The rate of change of momentum of an object is directly proportional to the applied unbalanced force and it takes place in the direction of the force." The mathematical expression is $F = ma$.
Since $a = \frac{v-u}{t}$, we have $F = m \frac{v-u}{t}$, which implies $Ft = mv - mu$.
If the external force $F = 0$, then $mv - mu = 0$, which means $mv = mu$, or $v = u$. This implies that if no external force acts on an object, its velocity remains constant (uniform motion). If $u = 0$, then $v = 0$, meaning the object remains at rest. This is the definition of Newton's first law of motion.
$(b)$ Given: $u = 0 \, m \, s^{-1}$, $t = 0.5 \, s$, $g = 10 \, m \, s^{-2}$.
$(i)$ Using $v = u + gt$: $v = 0 + 10 \times 0.5 = 5 \, m \, s^{-1}$.
$(ii)$ Average speed = $\frac{v + u}{2} = \frac{5 + 0}{2} = 2.5 \, m \, s^{-1}$.
$(iii)$ Using $h = ut + \frac{1}{2}gt^2$: $h = 0 + \frac{1}{2} \times 10 \times (0.5)^2 = 5 \times 0.25 = 1.25 \, m$.