## Question 3:

### Part (i):

**Either:**

| Answer | Mark | Guidance |
|--------|------|----------|
| Both components of initial speed: Horiz $31\cos 20°\ (29.1)$, Vert $31\sin 20°\ (10.6)$ | B1 | No credit if sin-cos interchanged. Components may be found anywhere in the question |
| Time to goal $= \dfrac{50}{31\cos 20°} = 1.716\ldots\ \text{s}$ | M1 | Attempt to use horizontal distance $\div$ horizontal speed |
| | A1 | |
| $h = 31\times\sin 20°\times1.716 + 0.5\times(-9.8)\times(1.716)^2$ | M1 | Use of one (or more) formula(e) to find the required result(s) relating to vertical motion within a correct complete method. Finding the maximum height is not in itself a complete method |
| $h = 3.76\ \text{(m)}$ | A1 | Allow 3.74 or other answers that would round to 3.7 or 3.8 if they result from premature rounding |
| So the ball goes over the crossbar | E1 | Dependent on both M marks. Allow follow through from previous answer |

**Or:**

| Answer | Mark | Guidance |
|--------|------|----------|
| Both components of initial speed | B1 | May be found anywhere in the question. No credit if sin-cos interchanged |
| $h = 31\sin 20°\times t - 4.9t^2$ | M1 | |
| Substitute $h = 2.44 \Rightarrow t = (0.26\ \text{or})\ 1.90$ | A1 | If only 0.26 is given, award A0 |
| Substitute $t = 1.90$ in $x = 31\cos 20°\times t$ | M1 | Allow this mark for substituting $t = 0.26$ |
| $x = 55.4$ | A1 | Allow $x = 7.6$ following on from $t = 0.26$ |
| Since $55.4 > 50$ the ball goes over the crossbar | E1 | Dependent on both M marks. Allow FT from their value for 55.4 |

**Or:**

| Answer | Mark | Guidance |
|--------|------|----------|
| Both components of initial speed | B1 | May be found anywhere in the question. No credit if sin-cos interchanged |
| $h = 31\sin 20°\times t - 4.9t^2$ | M1 | |
| Substitute $h = 2.44 \Rightarrow t = (0.26\ \text{or})\ 1.90$ | A1 | |
| Time to goal $= \dfrac{50}{31\cos 20°} = 1.716\ldots\ \text{s}$ | M1 | Attempt to use horizontal distance $\div$ horizontal speed |
| | A1 | |
| Since $1.90 > 1.72$ the ball goes over the crossbar | E1 | Dependent on both M marks. Allow follow through from previous answer |

**Or:**

| Answer | Mark | Guidance |
|--------|------|----------|
| Use of the equation of the trajectory | M1 | |
| $y = x\tan 20° - \dfrac{9.8x^2}{2\times31^2\times\cos^2 20°}$ | A1 | Correct substitution of $\alpha = 20°$ |
| | A1 | Fully correct |
| Substituting $x = 50$ | M1 | |
| $\Rightarrow y = 3.76$ | A1 | |
| So the ball goes over the crossbar | E1 | Dependent on both M marks. Follow through from previous answer |

### Part (ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| Any one reasonable statement | B1 | **Accept:** The ground is horizontal; The ball is initially on the ground; Air resistance is negligible; Horizontal acceleration is zero; The ball does not swerve; There is no wind; The particle model is being used; The value of $g$ is 9.8. **Do not accept:** $g$ is constant |
| | **[1]** | |

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