## Question 2:

### Part (i):

| Answer | Mark | Guidance |
|--------|------|----------|
| Vertical component of initial velocity $= 20\sin 30°\ (= 10)$ | B1 | |
| Vertical motion $s = s_0 + ut + \frac{1}{2}at^2$ | M1 | Substitution required. Sign of $g$ must be correct. Condone no $s_0$ |
| When it hits the sea $0 = 75 + 10t - 5t^2$ | A1 | |
| $75 + 10\times5 - 5\times5^2 = 0$ As required | | Or equivalent, e.g. solving the quadratic equation |
| This is satisfied when $t = 5$ | E1 | |

**Alternative:**

| Answer | Mark | Guidance |
|--------|------|----------|
| Vertical component of initial velocity $= 20\sin 30°\ (= 10)$ | B1 | |
| Vertical motion $v = u + at$ | M1 | Complete method for finding $t = 5$ required |
| At the top $0 = 10 - 10t \Rightarrow t = 1$ | | |
| It takes another 1 second to reach the level of the cliff top | | |
| At that point its speed is $10\ \text{ms}^{-1}$ downwards | | |
| When it hits the sea $-75 = -10t - 5t^2$ | | Or equivalent finding the time (4 seconds) from the top (height 80 m) to hitting the sea |
| $t^2 + 2t - 15 = 0 \Rightarrow t = 3$ | A1 | |
| Total time $= 1 + 1 + 3 = 5$ seconds | E1 | |
| Horizontal motion $x = 20\times\cos 30°\times t$ | M1 | |
| $t = 5 \Rightarrow 86.6$ | | |
| It is 3.4 m from the ship so within 5 m | E1 | Condone 3.5 m |
| | **[6]** | |

### Part (ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| It is longer in the air so it goes further | B1 | Justification for travelling further is required for this mark |
| | **[1]** | |

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