### (i)
Horiz $21t = 60$ | M1 | Use of horizontal components and $a = 0$ or $s = vt - 0.5at^2$ with $v = 0$.

so $\frac{30}{7}$ s $(2.8571...)$ | A1 | Any form acceptable. Allow M1 A1 for answer seen WW. [If $s = ut + 0.5at^2$ and $u = 0$ used without justification award M1 A0] [If $u = 28$ assumed to find time then award SC1]

either $0 = u - 9.8 \times \frac{30}{7}$ | M1 | Use of $v = u + at$ (or $v^3 = u^2 + 2as$) with $v = 0$. or Use of $v = u + at$ with $v = -u$ and appropriate $t$. or Use of $s = ut + 0.5at^2$ with $s = 40$ and appropriate $t$. Condone sign errors and, where appropriate, $u \leftrightarrow v$. Accept signs not clear but not errors.

or $-u = u - 9.8 \times (\frac{30}{7})$ | | 

or $40 = u \times \frac{30}{7} - 4.9 \times (\frac{30}{7})^2$ | | 

so $u = 28$ so $28$ m s$^{-1}$ | E1 | Enough working must be given for 28 to be properly shown. [NB $u = 28$ may be found first and used to find time]
| Sub: 4

### (ii)
$y = 28t - 0.5 \times 9.8t^2$ | E1 | Clear & convincing use of $g = -9.8$ in $s = ut + 0.5at^2$ or $s = vt - 0.5at^2$. NB: AG
| Sub: 1

### (iii)
Start from same height with same (zero) vertical speed at same time, same acceleration | E1 | For two of these reasons

Distance apart is $0.75 \times 21t = 15.75t$ | M1, A1 | $0.75 \times 21t$ seen or $21t$ and $5.25t$ both seen with intention to subtract. Need simplification - LHS alone insufficient. CWO.
| Sub: 3

### (iv)(A)
either Time is $\frac{30}{7}$ s by symmetry so $15.75 \times \frac{30}{7} = 45$ so $45$ m | B1, B1 | Symmetry or uvast. FT their (iii) with $t = \frac{30}{7}$

or Hit ground at same time. By symmetry one travels 60 m so the other travels 15 m in this time ($\frac{1}{2}$ speed) so 45 m. | B1 | 

| B1 | [SC1 if 90 m seen]
| Sub: 2

### (iv)(B)
[SC1 if either and or methods mixed to give $\pm30 = 28t - 4.9t^2$ or $\pm10 = 4.9t^2$]

either Time to fall is $40 = 10 = 0.5 \times 9.8 \times t^2$ | M1 | Considering time from explosion with $u = 0$. Condone sign errors. LHS. Allow $\pm30$

$t = 2.47435...$ need $15.75 \times 2.47435...= 38.971...$ so $39.0$ (3sf) | A1, A1, A1 | All correct. cao

or Need time so $10 = 28t - 4.9t^2$ | M1 | Equating $28t - 4.9t^2 = \pm10$. Dep. Attempt to solve quadratic by a method that could give two roots.

$4.9t^2 - 28t + 10 = 0$ | M1' | 

so $t = \frac{28 \pm \sqrt{784 - 4(4.9)(10)}}{9.8}$ so $0.382784...$ or $5.33150...$ | A1 | Larger root correct to at least 2 s. f. Both method marks may be implied from two correct roots alone (to at least 1 s. f.). [SC1 for either root seen WW]

Time required is $5.33150... - \frac{30}{7} = 2.47435...$ need $15.75 \times 2.47435...= 38.971...$ so $39.0$ (3sf) | M1, F1 | FT their (iii) only.
| Sub: 5

### (v)
Horiz ($x = $) $21t$ | B1 | 

Elim $t$ between $x = 21t$ and $y = 28t - 4.9t^2$ | M1 | Intention must be clear, with some attempt made. $t$ completely and correctly eliminated from their expression for $x$ and correct $y$. Only accept wrong notation if subsequently explicitly given correct value e.g. $\frac{x}{21}$ seen as $\frac{x}{21}$.

so $y = 28(\frac{x}{21}) - 4.9(\frac{x}{21})^2$ | A1 | Some simplification must be shown.

so $y = \frac{4x}{3} - \frac{0.1x^2}{9} = \frac{1}{90}(120x - x^2)$ | E1 | [SC2 for 3 points shown to be on the curve. Award more only if it is made clear that (a) trajectory is a parabola (b) 3 points define a parabola]
| Sub: 4

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## Total Marks: 60