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

Since the constant acceleration vector \( \vec{a} \) is not parallel or antiparallel to the initial velocity vector \( \vec{v} \) (here \( \vec{v} \cdot \vec{a} = 0 \)), the trajectory of the particle is parabolic.

Using third equation of motion: \(v^2 = u^2 - 2g(H/2) \Rightarrow 10^2 = u^2 - gH\). At max height, \(0 = u^2 - 2gH \Rightarrow u^2 = 2gH\). Thus, \(100 = 2gH - gH = gH \Rightarrow H = 10\text{ m}\).

Both maximum height \(H = \frac{u^2 \sin^2\theta}{2g}\) and time of flight \(T = \frac{2u \sin\theta}{g}\) are proportional to velocity \(u\). Since particle B has a larger velocity, both values are greater for B.

A projectile experiences constant acceleration (g) but follows a parabolic path. Also, a ball thrown vertically upwards under gravity has constant acceleration but reverses its direction of motion, making both statements false.

When velocity and acceleration have opposite signs, the speed decreases (retardation). After stopping, the positive acceleration will move the particle in the positive direction, which can result in positive displacement, so R is false.

Assertion (A) is false. Projectile motion has constant acceleration (\(\vec{g}\)) but follows a parabolic path, not a straight line.

Reason (R) is false. A particle can change direction even with constant acceleration (e.g., projectile motion, or an object slowing down and reversing direction).

Thus, both (A) and (R) are false.