(i) The greatest speed of the particle occurs at the lowest point M. Using conservation of energy, the loss of potential energy (PE) is converted into kinetic energy (KE):

\(Loss of PE = KE gain\)

\(mgh = \frac{1}{2} mv^2\)

\(0.5 \times 9.8 \times 2.45 = \frac{1}{2} \times 0.5 \times v^2\)

\(v^2 = 24.5\)

\(v = 7 \text{ m/s}\)

(ii) The kinetic energy of the particle at N is calculated by considering the height difference from L to N:

\(\text{KE at } N = 0.5 \times g \times (2.45 - 1.2)\)

\(= 0.5 \times 9.8 \times 1.25\)

\(= 6.25 \text{ J}\)

(iii) To find the least value of v for which the particle will reach L, use the kinetic energy at N:

\(\frac{1}{2} mv^2 = 6.25\)

\(\frac{1}{2} \times 0.5 \times v^2 = 6.25\)

\(v^2 = 25\)

\(v = 5 \text{ m/s}\)