A farmer models the height, \(h\) metres, of water from a rotating sprinkler, \(t\) seconds after starting, as \(h=6t-t^{2}\) for \(0 \leqslant t \leqslant ...
Question 1 Report
A farmer models the height, \(h\) metres, of water from a rotating sprinkler, \(t\) seconds after starting, as \(h=6t-t^{2}\) for \(0 \leqslant t \leqslant 6\). The graph is shown.
- Factorise \(6t-t^{2}\). (1)
- Write down the two values of \(t\) for which \(h=0\). (1)
- Find the maximum height reached, by completing the square, and the value of \(t\) at which it occurs. (3)
- Use the graph to estimate for how many seconds the height is above \(5\) m. (2)
Answer Details
(a) Both terms of \(6t-t^{2}\) share a common factor of \(t\), so \(6t-t^{2}=t(6-t)\). [1 mark]
(b) The height is zero when either factor is zero: \(t=0\) (the moment the sprinkler starts) or \(t=6\) (when it returns to the ground). [1 mark]
(c) Completing the square on \(h=6t-t^{2}=-(t^{2}-6t)=-[(t-3)^{2}-9]=9-(t-3)^{2}\). Since \(-(t-3)^{2}\) is never positive, its largest possible value is \(0\), which occurs when \(t=3\), giving a maximum height of \(9\) m at \(t=3\) seconds. [3 marks]
(d) Setting \(h=5\): \(6t-t^{2}=5\), so \(t^{2}-6t+5=0\), which factorises as \((t-1)(t-5)=0\), giving \(t=1\) and \(t=5\) seconds (matching where the graph crosses the \(h=5\) line). The height is above \(5\) m between these times, a duration of \(5-1=4\) seconds. [2 marks]
The vertex form from part (c) reveals the peak directly, while solving \(h=5\) separately in part (d) is needed because the vertex form does not immediately give the two times either side of the peak where a specific non-zero height is reached.
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