Horizontal range is given by \(R = \frac{u^2 \sin(2\theta)}{g}\). For complementary projection angles \(\theta\) and \(90^\circ - \theta\) (like \(30^\circ\) and \(60^\circ\)), the horizontal ranges are equal. Thus, the ratio of range is \(1 : 1\).

Maximum height \(H = \frac{u^2\sin^2\theta}{2g}\) and range \(R = \frac{2u^2\sin\theta\cos\theta}{g}\). Equating the two yields \(\frac{\sin^2\theta}{2} = 2\sin\theta\cos\theta\), which simplifies to \(\tan\theta = 4\).

Since the inclination equals the angle of repose, \( \mu = \tan\theta \). The acceleration down the incline during upward motion is \( a = g \sin\theta + \mu g \cos\theta = 2g \sin\theta \). Using \( v^2 = 2as \), we get \( v = 2\sqrt{gs \sin\theta} \).

For equal distances covered in two halves of a journey, average speed is \(v_{\text{avg}} = \frac{2v_1v_2}{v_1+v_2}\). Here, \(v_{\text{avg}} = \frac{2 \times 20 \times 40}{20+40} = \frac{80}{3}\text{ ms}^{-1}\).

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 \(H = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 9 = 45\text{ m}\).

Since acceleration is uniform, the distance is given by the formula \(s = \frac{u+v}{2} \times t\). Substituting \(u=20\text{ m/s}\), \(v=60\text{ m/s}\), and \(t=5\text{ s}\), we get \(s = \frac{20+60}{2} \times 5 = 200\text{ m}\).

We require \(H = \frac{R}{2}\). Using the formulae \(H = \frac{u^2 \sin^2 \theta}{2g}\) and \(R = \frac{u^2 \sin 2\theta}{g}\), we get \(\frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin\theta\cos\theta}{g} ⇒ \tan\theta = 2 ⇒ \theta = \tan^{-1}(2)\).

Boat speed in still water \(v_b = 6\text{ km/h}\), river speed \(v_r = 2\text{ km/h}\). Upstream speed \(v_u = v_b - v_r = 4\text{ km/h}\). Downstream speed \(v_d = v_b + v_r = 8\text{ km/h}\). Time upstream \(t_u = 2\text{ km} / 4\text{ km/h} = 0.5\) hours. Time downstream \(t_d = 2\text{ km} / 8\text{ km/h} = 0.25\) hours. Total time = \(0.5 + 0.25 = 0.75\) hours = \(45\) minutes.

Position of B: \(y_B(t) = H - (1/2)gt^2\). Meeting at \(H/2\): \(H/2 = H - (1/2)gt^2 \Rightarrow (1/2)gt^2 = H/2 \Rightarrow t = \sqrt{H/g}\). Position of A: \(y_A(t) = v_0t - (1/2)gt^2\). At meeting: \(H/2 = v_0t - H/2 \Rightarrow H = v_0t\). Substitute \(t\): \(H = v_0 \sqrt{H/g}\). Squaring both sides gives \(H^2 = v_0^2 (H/g) \Rightarrow v_0^2 = gH\).

Let the length of the escalator be \(L\). Speed of person is \(v_p = L/45\). Speed of escalator is \(v_e = L/60\). When both move, their effective speed is \(v_{eff} = v_p + v_e = L/45 + L/60 = (4L+3L)/180 = 7L/180\). The time taken \(T = L/v_{eff} = L / (7L/180) = 180/7 \approx 25.71\) s.