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.

Let the bus's position be \(x_b = (1/2)at^2 = t^2\). The cyclist's position (starting 96m behind) is \(x_c = 20t - 96\). For meeting, \(t^2 = 20t - 96 \Rightarrow t^2 - 20t + 96 = 0\). Factoring gives \((t-8)(t-12) = 0\), so \(t=8\)s or \(t=12\)s. At \(t=8\)s, cyclist is faster than bus. At \(t=12\)s, bus is faster, so it overtakes.

In still water, total distance is 20 km in 2 hours, so boat speed \(v_b = 10\text{ km/h}\). Water velocity \(v_w = 5\text{ km/h}\). Upstream speed \(v_b - v_w = 5\text{ km/h}\). Time upstream = \(10/5 = 2\) hours. Downstream speed \(v_b + v_w = 15\text{ km/h}\). Time downstream = \(10/15 = 2/3\) hours. Total time = \(2 + 2/3 = 8/3\) hours = \(160\) minutes.

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.

Both bodies are released simultaneously and fall under gravity. Their acceleration is identical (\(g\)). Since their initial relative velocity is zero and relative acceleration is zero, their relative distance remains constant. Thus, after \(2\) s, the relative distance is still \(9.8\) m.

Let \(y_A = 20t - 5t^2\) and \(y_B = 80 - 20t - 5t^2\). Equating them for collision: \(20t - 5t^2 = 80 - 20t - 5t^2\). This simplifies to \(40t = 80\), so \(t = 2\) s. The collision height is \(y_A(2) = 20(2) - 5(2)^2 = 40 - 20 = 20\) m above the ground.

First walnut (W1): Time to fall 5m from rest is \(t_1 = \sqrt{2s/g} = \sqrt{2 \cdot 5/10} = 1\) s. Total time for W1 to reach ground from 20m is \(t_{total} = \sqrt{2 \cdot 20/10} = 2\) s. Second walnut (W2) must fall 15m in \(t_{W2} = t_{total} - t_1 = 2-1 = 1\) s. Using \(s = v_0t + (1/2)gt^2\): \(15 = v_0(1) + (1/2)(10)(1)^2 \Rightarrow 15 = v_0 + 5 \Rightarrow v_0 = 10\text{ m/s}\).

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 velocities be \(v_1\) and \(v_2\). Distance to cross is \(100 + 100 = 200\text{ m}\). For same direction: \(v_1 - v_2 = 200/20 = 10\text{ m/s}\). For opposite direction: \(v_1 + v_2 = 200/10 = 20\text{ m/s}\). Solving these gives \(v_1 = 15\text{ m/s}\) and \(v_2 = 5\text{ m/s}\).

Range is given by \(R = \frac{u^2 \sin(2\theta)}{g}\). For \(\theta = 15^\circ\), \(R_1 = \frac{u^2 \sin(30^\circ)}{g} = 1.5\text{ km}\) which implies \(\frac{u^2}{g} = 3.0\text{ km}\). For \(\theta = 45^\circ\), \(R_2 = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g} = 3\text{ km}\).