# Question 7:

## Part (i):

| Working | Mark | Guidance |
|---------|------|----------|
| $v\sin50°$ | B1 | initial vertical component |
| $0 = v^2\sin^250° - 2\times9.8\times13$ (must be 13) | M1 | or $mx9.8x13 = \frac{1}{2}m(v\sin50°)^2$ |
| $v = 20.8$ ms$^{-1}$ | A1 | 3 marks total; sin/cos mix ok for above M1 |

## Part (ii):

| Working | Mark | Guidance |
|---------|------|----------|
| $45 = v\cos50° \cdot t$ | M1 | see alternative below |
| $t = 3.36\ \checkmark$ their $v$ ($3.13$ for $v=22.4$) | A1$\checkmark$ | other methods include other $t_s$ |
| $s = v\sin50° \times t - \frac{1}{2} \times 9.8 \times t^2$ | M1 | ignore height adjustments |
| $s = -1.6$ to $-2.0$ inclusive $(-1.68)$ | A1, A1 | can be their $v$ and their $t$; can be implied from next A1 |
| height above ground $= 0.320$ m | A1 | 6 marks total |

## Part (iii):

| Working | Mark | Guidance |
|---------|------|----------|
| $v_v = v\sin50° - 9.8xt$ | M1 | or $v_v^2 = 2g(15\text{-their ans to ii})$ |
| $v_v = -17.0\ \checkmark$ their $v,t$ ($-13.5$ for $22.4$) | A1$\checkmark$ | $\checkmark$ above for $v_v$ |
| speed $= \sqrt{v_v^2 + (v\cos50°)^2}$ | M1 | or $\frac{1}{2}mv^2 - mgx1.68 =$ |
| speed $= 21.6$ ms$^{-1}\ \checkmark$ their $v$ and $v_v$ ($19.7$ for $v=22.4$) | A1$\checkmark$ | 4 marks total; $\frac{1}{2}mx20.8^2$ (4 marks) M1/A1$\checkmark$ s,v/M1 solve/A1$\checkmark$; 13 marks total |

### Alternative Part (ii) — 1st 5 marks:

| Working | Mark | Guidance |
|---------|------|----------|
| $y = x\tan\theta - \frac{gx^2}{2v^2\cos^2\theta}$ | B1 | substitute $v$ and $50°$ and $x=45$ |
| $y = 45\tan50° - 9.8 \cdot 45^2 / 2 \cdot v^2\cos^250°$ | M1 | |
| calculate $y$ | A1, M1 | can be their $v$ |
| $y = -1.6$ to $-2.0$ inclusive | A1 | should be $-1.68$ |

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