June 2023 Paper 1 Q15
15 A projectile is launched from a point on level ground with an initial velocity \(u\) at an angle \(\theta\) above the horizontal.
| Scheme | Marks | AO |
|---|---|---|
| vertical motion with \(u\sin\theta,\ a = -g,\ s = 0\) \(0 = u\sin\theta\, t - \frac{1}{2}gt^2\) \(t = 0,\ \dfrac{2u\sin\theta}{g}\) | M1 | 2.1 |
| horizontal distance \(R\) given by \(t = \dfrac{2u\sin\theta}{g},\ R = u\cos\theta \times \dfrac{2u\sin\theta}{g}\) | M1 | 2.1 |
| \(R = \dfrac{2u^2\sin\theta\cos\theta}{g}\) | A1 | 2.1 |
| [3] |
Notes
M1: suvat equations used with \(u\sin\theta\) leading to an expression for the time of flight. Do not allow for time to the top unless subsequently doubled.
M1: horizontal motion with \(u\cos\theta,\ a = 0\) and their expression for \(t\) soi
Allow sin/cos interchange if consistent with their vertical equation
A1: Convincing argument AG
Note M0M1A0 for sin/cos interchange
Alternative method
| Scheme | Marks |
|---|---|
| Substitute \(t = \dfrac{x}{u\cos\theta}\) to form equation of the trajectory \(y = u\sin\theta \times \dfrac{x}{u\cos\theta} - \dfrac{1}{2}g\left(\dfrac{x}{u\cos\theta}\right)^2\) | M1 |
| Equate \(y\) to zero and attempt to rearrange | M1 |
| \(R = \dfrac{2u^2\sin\theta\cos\theta}{g}\) | A1 |
M1: Allow equivalent formula quoted
M1: Allow sin/cos interchange if consistent with their horizontal equation
A1: Convincing argument AG
| Scheme | Marks | AO |
|---|---|---|
| Max height \(H\) when \(v_y = 0\) \(0 = (u\sin\theta)^2 - 2gH\) | M1 | 3.1b |
| \(H = \dfrac{u^2\sin^2\theta}{2g}\) | A1 | 1.1b |
| Max height exceeds range when \(\dfrac{u^2\sin^2\theta}{2g} > \dfrac{2u^2\sin\theta\cos\theta}{g}\) | M1 | 1.1a |
| \(\tan\theta > 4\) | M1 | 1.1a |
| \(76.0^\circ < \theta\ [< 90^\circ]\) | A1 | 1.1b |
| [5] |
Notes
M1: suvat equation(s) with \(v_y = 0\) leading to an equation for \(H\) not involving \(t\). Allow sin/cos interchange if consistent with their (a)
A1: correct expression for \(H\)
M1: Compares their \(H\) with given \(R\)
Allow = used to find boundary value
M1: simplifies the inequality to an inequality for \(\tan\theta\) (or equation)
A1: must be an inequality for \(\theta\). Do not penalise for omission of \(90^\circ\)