2. The height of a small glider is described by the function $h(x)=2 x^{3}-17 x^{2}+40 x$ where $x$ is the time in minutes and $h(x)$ is the height in hundreds of metres. When will the height be 2500 $\mathrm{m}$ ? Provide a full a/gebraic solution. [5 marks]

Step 1 ：We are given the function \(h(x) = 2x^3 - 17x^2 + 40x\) which describes the height of a glider in hundreds of meters over time in minutes. We are asked to find the time at which the height of the glider is 2500m. Since the height is given in hundreds of meters, we convert 2500m to 25 hundreds of meters.

Step 2 ：We set \(h(x)\) equal to 25 and solve for \(x\): \(2x^3 - 17x^2 + 40x = 25\). The solutions to this equation are \(x = 1\), \(x = \frac{5}{2}\), and \(x = 5\).

Step 3 ：However, these are the times at which the glider reaches a height of 2500m. Since the glider cannot be at the same height at different times, we need to determine at which of these times the glider is ascending. We do this by finding the derivative of the height function and checking the sign of the derivative at the given times.

Step 4 ：The derivative of the height function is \(h'(x) = 6x^2 - 34x + 40\). The values of the derivative at the times \(x = 1\), \(x = \frac{5}{2}\), and \(x = 5\) are 12, -7.5, and 20 respectively.

Step 5 ：The negative derivative at \(x = \frac{5}{2}\) indicates that the glider is descending at this time, so this solution is not valid. The positive derivatives at \(x = 1\) and \(x = 5\) indicate that the glider is ascending at these times, so these solutions are valid.

Step 6 ：However, since the glider cannot be at the same height at different times, we need to choose the earliest time, which is \(x = 1\).

Step 7 ：Final Answer: The glider will be at a height of 2500m at \(x = 1\) minute. So, the final answer is \(\boxed{1}\).