# Question 6:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Both components of initial speed: Horiz $31\cos20°\ (29.1)$, Vert $31\sin20°\ (10.6)$ | B1 | No credit if sin-cos interchanged. Components may be found anywhere in the question |
| Time to goal $= \dfrac{50}{31\cos20°} = 1.716\ldots\ \text{s}$ | M1, A1 | Attempt to use horizontal distance $\div$ horizontal speed |
| $h = 31\times\sin20°\times1.716 + 0.5\times(-9.8)\times(1.716)^2$ | M1 | Use of formula(e) to find required result(s) relating to vertical motion within a correct complete method. Finding maximum height is not itself a complete method |
| $h = 3.76\ \text{(m)}$ | A1 | Allow 3.74 or other answers rounding to 3.7 or 3.8 from premature rounding |
| So the ball goes over the crossbar | E1 | Dependent on both M marks. Allow follow through |

**Or** (trajectory equation method):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of equation of trajectory | M1 | |
| $y = x\tan20° - \dfrac{9.8x^2}{2\times31^2\times\cos^220°}$ | A1, A1 | Correct substitution of $\alpha=20°$; fully correct |
| Substituting $x=50$ | M1 | |
| $\Rightarrow y = 3.76$ | A1 | |
| So the ball goes over the crossbar | E1 | Dependent on both M marks |

## Part (ii)
| Answer | Marks | 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 |