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 \(\theta = 45^\circ\), \(R_2 = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g} = 3\text{ km}\).

The particle comes to rest when \(v = 0\), giving \(4t - 2 = 0 ⇒ t = 0.5\text{ s}\). Integrating velocity to find position, \(x = int_0^t (4t - 2)dt = 2t^2 - 2t\). At \(t = 0.5\text{ s}\), \(x = 2(0.5)^2 - 2(0.5) = -0.5\text{ m}\).

The distance covered in the \(n\)-th second from rest is given by \(S_n = frac{a}{2}(2n - 1)\). For the first second (\(n=1\)), \(S_1 = frac{a}{2}\), and for the third second (\(n=3\)), \(S_3 = frac{5a}{2}\). The ratio of \(S_1 : S_3\) is \(1 : 5\).

The distance in the \(n^{\text{th}}\) second is \(S_n = u + \frac{a}{2}(2n-1)\). Since \(u=0\), \(S_n \propto 2n-1\). Thus, \(\frac{S_n}{S_{n+1}} = \frac{2n-1}{2(n+1)-1} = \frac{2n-1}{2n+1}\).

At \(t = 4\text{ s}\), the horizontal velocity is \(v_x = u + at = 0 + 5 \times 4 = 20\text{ m/s}\) and remains constant. In the vertical direction, the ball accelerates due to gravity for \(2\text{ s}\) (from \(t = 4\text{ s}\) to \(6\text{ s}\)), so \(v_y = gt = 10 \times 2 = 20\text{ m/s}\). The final velocity is \(v = \sqrt{v_x^2 + v_y^2} = 20\sqrt{2}\text{ m/s}\), and the acceleration is only due to gravity (\(10\text{ m/s}^2\)).

Formula: For equal halves of distance, the average speed is the harmonic mean: \(v_{av} = \frac{2v_1v_2}{v_1+v_2}\). Putting values, \(v_{av} = \frac{2 \times 2 \times 5}{2+5} = \frac{20}{7}\text{ m/s}\).

Formula: Under uniform acceleration, the midpoint velocity is \(v_{mid} = \sqrt{\frac{v_1^2 + v_2^2}{2}}\). Thus, \(v_{mid} = \sqrt{\frac{100 + 400}{2}} = 5\sqrt{10}\text{ m/s}\).

Relative velocity is given by \(\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = (3\hat{i}+4\hat{j}) - (7\hat{i}-3\hat{j}) = -4\hat{i}+7\hat{j}\) m/s. Therefore, option D is correct.

For minimum crossing time, the swimmer must head perpendicular to the river bank. The net ground velocity is the vector sum of swimmer's velocity and river velocity: \(v_g = \sqrt{v_{\text{sw}}^2 + v_r^2} = \sqrt{8^2 + 4^2} = \sqrt{80} \approx 8.94\text{ m/s} \approx 9\text{ m/s}\).