The time taken by an object to fall from a height is given by the equation:

\[
t = \sqrt{\frac{2h}{g}}
\]

For mass \( m_a \) dropped from height \( a \), the time taken is:

\[
t_a = \sqrt{\frac{2a}{g}}
\]

For mass \( m_b \) dropped from height \( b \), the time taken is:

\[
t_b = \sqrt{\frac{2b}{g}}
\]

The ratio of time taken by the two bodies is:

\[
\frac{t_a}{t_b} = \frac{\sqrt{\frac{2a}{g}}}{\sqrt{\frac{2b}{g}}} = \sqrt{\frac{a}{b}}
\]

So, the ratio of the time taken is:

\[
\frac{t_a}{t_b} = \sqrt{\frac{a}{b}}
\]

\[ t =\frac{2u}{g}=\frac{2\times 30}{10}= 6 sec\]

Distance in last second is \(h_{\text{last}} = \frac{5}{9}H ⇒ \text{Distance in } (t-1) \text{ seconds is } \frac{4}{9}H\). Therefore, \(\frac{\frac{1}{2}g(t-1)^2}{\frac{1}{2}gt^2} = \frac{4}{9} ⇒ \frac{t-1}{t} = \frac{2}{3} ⇒ t = 3\text{ s}\). Height \(H = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 9 = 45\text{ m}\).

If the air resistance force \(F_{/text{air}}\) is the same for both bodies, the net force is \(mg - F_{/text{air}}\). The acceleration is \(a = g - F_{/text{air}}/m\). Since masses \(M\) and \(m\) are different, their accelerations will be different. Thus, they will not reach the earth simultaneously, and their accelerations will not be the same. Both assertion and reason are false.

Formula: \(v^2 = u^2 + 2gs\), \(g \approx 9.8 \text{ m/s}^2\).
Solution: (A) is false; for a fall of \(5 \text{ m}\) from rest, \(v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.9 \text{ m/s}\), not \(5 \text{ m/s}\). (R) is false; acceleration due to gravity is approximately \(9.8 \text{ m/s}^2\) or \(10 \text{ m/s}^2\), not \(5 \text{ m/s}^2\).

Assertion (A): Let \(\Delta t\) be the time interval. The distance between them \(D(t) = \frac{1}{2}gt^2 - \frac{1}{2}g(t-\Delta t)^2 = g t \Delta t - \frac{1}{2}g (\Delta t)^2\). This is a linear function of time \(t\). So (A) is true.

Reason (R): Both balls accelerate at \(g\). Thus, their relative acceleration is \(\vec{g} - \vec{g} = \vec{0}\). The first ball has velocity \(g\Delta t\) when the second is dropped, so the relative velocity is non-zero and constant. So (R) is true.
(R) correctly explains (A): constant non-zero relative velocity results in linear increase in relative distance.