Mechanics

Could anyone explain why you have to use those values for a and s for iii)?

Motion is divided into two parts: before P hits the ground, and after.

Initially Q is on the ground at 0.4m below the pulley. So we know the string must be more than 0.4m. Let this extra length be denoted as d so that total length is 0.4+d.

BEFORE THE GROUND HIT:

a is the acceleration of the two particles. They are connected with a taut string so they have the same acceleration. It is calculated as 5.88 ms^-2 in part (ii). Hence, P travels with this acceleration downward. How much does it travel? Initially it must be hanging at height 0.4-d above ground. So this is the distance it travels to hit the ground and stay there. That's the s value for this particle.

What happens to Q? When P travels (0.4-d)m down, Q travels the same amount upward. So it must be at distance 0.4-(0.4-d) = d below the pulley when P hits the ground.

AFTER THE GROUND HIT:

Once P hits the ground, the string is now slack and the only force acting on Q is its weight. Hence the acceleration has magnitude g. Q must travels this distance d to the pulley under this acceleration whereby it comes to rest.

When p hit the ground, would the acceleration for Q not be -g if Q is travelling up?

Yes, that is taken into account, but that is not clear from the way the answer here is written.

In v^2 = u^2 + 2as we have initial speed u and final speed v.

This is used for particle P to give

v^2 = 0^2 + 2(5.88)(0.4-d)

This v is the speed of P as it hits the ground.

This is also the same speed that Q has as the string becomes slack. So, this is actually the initial speed of Q in the second part of motion. The final speed of Q here is 0 since it comes to rest at the pulley.

Hence, using the same suvat formula for Q, we have:

0^2 = v^2 - 2gd

Or, as they write it, v^2 = 0^2 + 2gd