15 (a) At time $t$ hours after a high tide, the height, $h$ metres, of the tide and the velocity, $v$ knots, of the tidal flow can be modelled using the parametric equations $v = 4 - rac{(2t}{3} - 2)^2$ $h = 3 - 2rac{(t - 3)}{3}$ High tides and low tides occur alternately when the velocity of the tidal flow is zero - AQA - A-Level Maths Mechanics - Question 15 - 2019 - Paper 1

To find the first low tide, we need to set the velocity $v$ to zero:

$0 = 4 - \left(\frac{(2t}{3} - 2\right)^2$

Solving for $t$, we get:

$\left(\frac{(2t}{3} - 2\right)^2 = 4$

Taking the square root gives us two solutions. Calculating, we find:

$\frac{(2t)}{3} - 2 = 2$ or $\frac{(2t)}{3} - 2 = -2$.

This results in $t = 6$ hours as the first positive solution (the second solution is negative). Therefore, the first low tide after 2 am occurs at 8 am.

Using the time calculated from the previous step, we substitute $t = 6$ into the height equation:

$h = 3 - 2\frac{(6 - 3)}{3}$

This evaluates to:

$h = 3 - 2\cdot 1 = 3 - 2 = 1$

Thus, the height of the low tide is $3$ metres.