For an object in free fall from rest, the initial velocity is $v_0 = 0$ and the acceleration is $a=g$. The total distance $d$ fallen after a time $t$ is given by the kinematic equation:

$$d(t) = v_0 t + \frac{1}{2} a t^2 = \frac{1}{2} g t^2$$

[Q1] To find the distance fallen during the nth second, we calculate the difference between the total distance fallen after $n$ seconds and the total distance fallen after $(n-1)$ seconds. Let this distance be $\Delta d_n$.

$$\Delta d_n = d(n) - d(n-1)$$

Substituting the expression for $d(t)$:

$$\Delta d_n = \frac{1}{2} g n^2 - \frac{1}{2} g (n-1)^2$$

Factor out the common term $\frac{1}{2}g$:

$$\Delta d_n = \frac{1}{2} g [n^2 - (n-1)^2]$$

Expand the squared term:

$$\Delta d_n = \frac{1}{2} g [n^2 - (n^2 - 2n + 1)]$$ $$\Delta d_n = \frac{1}{2} g (2n - 1)$$

Now we can find the distance for each successive second:

• 1st second (n=1): $\Delta d_1 = \frac{1}{2} g (2(1) - 1) = \frac{1}{2} g (1)$
• 2nd second (n=2): $\Delta d_2 = \frac{1}{2} g (2(2) - 1) = \frac{1}{2} g (3)$
• 3rd second (n=3): $\Delta d_3 = \frac{1}{2} g (2(3) - 1) = \frac{1}{2} g (5)$
• nth second (n): $\Delta d_n = \frac{1}{2} g (2n - 1)$

The ratio of these distances is:

$$\Delta d_1 : \Delta d_2 : \Delta d_3 : \dots : \Delta d_n$$ $$\frac{1}{2} g (1) : \frac{1}{2} g (3) : \frac{1}{2} g (5) : \dots : \frac{1}{2} g (2n-1)$$

Canceling the common factor $\frac{1}{2}g$, we get the ratio of successive odd integers.