| Date | May 2017 | Marks available | 5 | Reference code | 17M.2.hl.TZ1.11 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Determine | Question number | 11 | Adapted from | N/A |
Question
Xavier, the parachutist, jumps out of a plane at a height of \(h\) metres above the ground. After free falling for 10 seconds his parachute opens. His velocity, \(v\,{\text{m}}{{\text{s}}^{ - 1}}\), \(t\) seconds after jumping from the plane, can be modelled by the function
\(v(t) = \left\{ {\begin{array}{*{20}{l}} {9.8t{\text{,}}}&{0 \leqslant t \leqslant 10} \\ {\frac{{98}}{{\sqrt {1 + {{(t - 10)}^2}} }},}&{t > 10} \end{array}} \right.\)
His velocity when he reaches the ground is \(2.8{\text{ m}}{{\text{s}}^{ - 1}}\).
Find his velocity when \(t = 15\).
Calculate the vertical distance Xavier travelled in the first 10 seconds.
Determine the value of \(h\).
Markscheme
\(v(15) = \frac{{98}}{{\sqrt {1 + {{(15 - 10)}^2}} }}\) (M1)
\(v(15) = 19.2{\text{ }}({\text{m}}{{\text{s}}^{ - 1}})\) A1
[2 marks]
\(\int\limits_0^{10} {9.8t\,{\text{d}}t} \) (M1)
\( = 490{\text{ }}({\text{m}})\) A1
[2 marks]
\(\frac{{98}}{{\sqrt {1 + {{(t - 10)}^2}} }} = 2.8\) (M1)
\(t = 44.985 \ldots {\text{ }}({\text{s}})\) A1
\(h = 490 + \int\limits_{10}^{44.9...} {\frac{{98}}{{\sqrt {1 + {{(t - 10)}^2}} }}{\text{d}}t} \) (M1)(A1)
\(h = 906{\text{ (m}})\) A1
[5 marks]
