## Question 3:

### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Speed $= \sqrt{(-5)^2 + 0^2 + (-10)^2}$ | M1 | For use of Pythagoras. Accept $\sqrt{5^2 + 10^2}$ |
| $= 11.2\ \text{ms}^{-1}$ (11.18) | A1 | Accept $\sqrt{125}$ or $5\sqrt{5}$ |
| $\tan\theta = \frac{5}{10}$ | M1 | Complete method for correct angle; may use $\sin\theta = \frac{5}{11.2}$, $\cos\theta = \frac{10}{11.2}$ |
| $\theta = 26.6°$ | A1 | Allow $153.4°$, $206.6°$ |
| **[4]** | | |

### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}0\\0\\-980\end{pmatrix}$ her weight | B1 | The descriptions should be linked to the forces, either by the layout of the answer or by suitable text |
| $\begin{pmatrix}0\\0\\880\end{pmatrix}$ resistance to her vertical motion | B1 | Accept "Air resistance", "Arms stretched out" and similar statements. Condone mention of a parachute |
| $\begin{pmatrix}50\\-20\\0\end{pmatrix}$ force from the power unit | B1 | |
| **[3]** | | |

### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resultant force $= \begin{pmatrix}50\\-20\\-100\end{pmatrix}$ | B1 | May be implied |
| Acceleration $= \begin{pmatrix}0.5\\-0.2\\-1\end{pmatrix}$ | B1 | Newton's 2nd Law |
| Magnitude $= \sqrt{0.5^2 + (-0.2)^2 + 1^2} = 1.1357...$ so $1.14$ to 3 s.f. | B1 | Answer given. Allow FT from sign errors. Accept $|\mathbf{F}| \div 100$ |
| **[3]** | | |

### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ | M1 | FT their $\mathbf{a}$ for the first 5 marks of this part. Vectors must be seen or implied. Accept valid integration |
| $\mathbf{v} = \begin{pmatrix}-5\\0\\-10\end{pmatrix} + \begin{pmatrix}0.5\\-0.2\\-1\end{pmatrix}t$ | A1 | |
| $\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ | M1 | Vectors must be seen or implied. Accept valid integration. Condone no $\mathbf{r}_0$ for this M mark |
| $\mathbf{r} = \begin{pmatrix}-75\\90\\750\end{pmatrix} + \begin{pmatrix}-5\\0\\-10\end{pmatrix}t + \frac{1}{2}\begin{pmatrix}0.5\\-0.2\\-1\end{pmatrix}t^2$ | A1 | |
| When $t = 30$: $\mathbf{r} = \begin{pmatrix}-75-150+225\\90+0-90\\750-300-450\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$, as required | M1, E1 | Vectors must be seen or implied. CAO |
| **[6]** | | SC1 to replace first 4 marks if acceleration taken to be $\mathbf{g}$ but answer is otherwise correct |

### Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t = 30$, $\mathbf{v} = \begin{pmatrix}10\\-6\\-40\end{pmatrix}$ | M1 | There must be an attempt to work out at least the vertical component of the velocity at $t = 30$. This mark is not dependent on a correct answer |
| The vertical component of the velocity is too fast for a safe landing | A1 | Accept an argument based on speed derived from a vector |
| **[2]** | | |

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