| Date | May 2017 | Marks available | 2 | Reference code | 17M.2.hl.TZ1.11 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Find | 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, vms−1, t seconds after jumping from the plane, can be modelled by the function
v(t)={9.8t,0⩽
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]
