# Question 10:

## Part (a):
| $0^2 = (U\sin\alpha)^2 - 2g \times (3-2)$ | M1 | 3.3 | Any complete method for equation in $U$, $g$, $\alpha$ only |
|---|---|---|---|
| $U^2 = \frac{2g}{\sin^2\alpha}$ | A1* | 2.2a | Given answer |

## Part (b):
| $20 = Ut\cos\alpha$ | M1, A1 | 3.4, 1.1b | Using horizontal motion $s = ut$ |
| $-\frac{5}{4} = Ut\sin\alpha - \frac{1}{2}gt^2$ | M1, A1 | 3.4, 1.1b | Using vertical motion $s = ut + \frac{1}{2}at^2$; M0 if $s = \pm2$ or $\pm3$ used, allow $s = \pm1.25$, $\pm0.75$, $\pm2.25$, $\pm2.75$ |
| $-\frac{5}{4} = U\sin\alpha\left(\frac{20}{U\cos\alpha}\right) - \frac{1}{2}g\left(\frac{20}{U\cos\alpha}\right)^2$ | M1(I) | 3.1b | Sub for $t$ using $20 = Ut\cos\alpha$ |
| Sub for $U^2$ using part (a) | M1(II) | 3.1b | |
| $-\frac{5}{4} = 20\tan\alpha - 100\tan^2\alpha$ | A1(I) | 1.1b | Correct quadratic in $\tan\alpha$ or $\cot\alpha$ |
| $(4\tan\alpha - 1)(100\tan\alpha + 5) = 0$ | M1(III) | 1.1b | Solve 3-term quadratic by factorisation, formula, or calculator |
| $\tan\alpha = \frac{1}{4} \Rightarrow \alpha = 14°$ or better | A1(II) | 2.2a | No restriction on accuracy since $g$'s cancel |

### Alternative method for (b):
| Sub $U\sin\alpha$ to give equation in $t$ only: $-\frac{5}{4} = \sqrt{2g}\,t - \frac{1}{2}gt^2$ | M1(II), A1(I) | | |
|---|---|---|---|
| Solve for $t$: $t = \frac{5}{\sqrt{2g}}$ or $1.1$ or $1.13$ | M1(III), M1(I) | | |
| Use $20 = Ut\cos\alpha \Rightarrow \alpha = 14°$ or better | A1(II) | | |

## Part (c):
| The target will have dimensions so there would be a range of possible values of $\alpha$; **or** there will be air resistance; **or** the ball will have dimensions; **or** wind effects; **or** spin of the ball | B1 | 3.5b | B0 if any incorrect extras, but ignore extra consequences |

## Part (d):
| Find $U$ using their $\alpha$, e.g. $U = \sqrt{\frac{2g}{\sin^2\alpha}}$ | M1 | 3.1b | |
|---|---|---|---|
| Use $20 = Ut\cos\alpha$ (or vertical motion equation) | A1 M1 | 1.1b | Treat as M1 |
| $t = \frac{5}{\sqrt{2g}}$ or $1.1$ or $1.13$ | B1 A1 | 1.1b | Treat as A1; $t = 1.1$ or $1.10$ since depends on $g = 9.8$ |

### Alternative for (d):
| $A$ to top: $s = vt - \frac{1}{2}at^2$ giving $t_1 = \sqrt{\frac{2}{g}}$; top to $T$: $s = ut + \frac{1}{2}at^2$ giving $t_2 = \frac{3}{\sqrt{2g}}$ | M1, A1 M1 | 3.1b, 1.1b | |
|---|---|---|---|
| Total time $= \sqrt{\frac{2}{g}} + \frac{3}{\sqrt{2g}} = \frac{5}{\sqrt{2g}} = 1.1$ or $1.13$ s | B1 A1 | 1.1b | |

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