# Question 5:

## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = \sqrt{10^2 + 12^2} = 15.62\ldots$ | B1 | Accept any accuracy 2 s.f. or better |
| $\theta = \arctan\left(\frac{12}{10}\right) = 50.1944\ldots$ so $50.2°$ (3 s.f.) | M1 | Accept $\arctan\left(\frac{10}{12}\right)$. Or **their** $15.62\cos\theta = 10$ or **their** $15.62\sin\theta = 12$ |
| | A1 | [FT **their** 15.62 if used]. [If $\theta$ found first M1 A1 for $\theta$, F1 for $u$]. [If B0 M0 SC1 for both $u\cos\theta = 10$ and $u\sin\theta = 12$ seen] |

## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| vert: $12t - 0.5 \times 10t^2 + 9$ | M1 | Use of $s = ut + 0.5at^2$, $a = \pm9.8$ or $\pm10$ and $u=12$ or 15.62. Condone $-9 = 12t - 0.5 \times 10t^2$, condone $y = 9 + 12t - 0.5 \times 10t^2$. Condone $g$ |
| $= 12t - 5t^2 + 9$ (AG) | A1, E1 | All correct with origin of $u=12$ clear; accept 9 omitted. Reason for 9 given. Must be clear unless $y = s_0 + \ldots$ used |
| horiz: $10t$ | B1 | |

## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = 12^2 - 20s$ | M1 | Use of $v^2 = u^2 + 2as$ or equiv with $u=12$, $v=0$. Condone $u \leftrightarrow v$ |
| $s = 7.2$ so $7.2$ m | A1 | From CWO. Accept 16.2 |

## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| We require $0 = 12t - 5t^2 + 9$ | M1 | Use of $y$ equated to 0 |
| Solve for $t$ | M1 | Attempt to solve a 3 term quadratic |
| the $+$ve root is 3 | A1 | Accept no reference to other root. cao |
| range is $30$ m | F1 | FT root and **their** $x$. [If range split up M1 all parts considered; M1 valid method for each part; A1 final phase correct; A1] |

## Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Horiz displacement of B: $20\cos 60t = 10t$ | B1 | Condone unsimplified expression. Award for $20\cos60 = 10$ |
| Comparison with horiz displacement of A | E1 | Comparison clear, must show $10t$ for each or explain |

## Part (vi)
| Answer/Working | Mark | Guidance |
|---|---|---|
| vertical height is $20\sin 60t - 0.5 \times 10t^2 = 10\sqrt{3}t - 5t^2$ (AG) | A1 | Clearly shown. Accept decimal equivalence for $10\sqrt{3}$ (at least 3 s.f.). Accept $-5t^2$ and $20\sin60 = 10\sqrt{3}$ not explained |

## Part (vii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Need $10\sqrt{3}t - 5t^2 = 12t - 5t^2 + 9$ | M1 | Equating the **given** expressions |
| $\Rightarrow t = \frac{9}{10\sqrt{3} - 12}$ | A1 | Expression for $t$ obtained in any form |
| $t = 1.6915\ldots$ so $1.7$ s (2 s.f.) (AG) | E1 | Clearly shown. Accept 3 s.f. or better as evidence. Award M1 A1 E0 for 1.7 sub in each ht |

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