June 2025 Paper 2 Q14

AQACurrent spec4 marksProjectiles

14 In this question use \(g = 9.8\ \text{m s}^{-2}\)

An arrow is projected from a point \(P\) which is at a height of 2.5 metres above the horizontal ground.

The arrow has an initial velocity of \(\begin{bmatrix}40\\25\end{bmatrix}\ \text{m s}^{-1}\)

The arrow lands on the horizontal ground at a point \(Q\)

The path of the arrow is shown in the diagram.

The arrow is projected from P, 2.5 m above horizontal ground, with velocity (40, 25) m/s, follows a parabolic path and lands at Q on the ground

Find the speed of the arrow at point \(Q\) [4 marks]

Mark scheme

Scheme	Marks	AO
Substitutes \(u = 25,\ a = \pm 9.8\) and \(s = \pm 2.5\) into \(v^2 = u^2 + 2as\) PI \(\begin{bmatrix}k\\674\end{bmatrix}\) or \(\begin{bmatrix}\sqrt{k}\\\sqrt{674}\end{bmatrix}\) Or Substitutes \(u = 25,\ a = -9.8\) and \(s = -2.5\) into \(s = ut + \frac{1}{2}at^2\) to find \(t\) and substitutes their \(t\) into \(v = u + at\)	M1	3.3
Obtains correct \(v^2 = 674\) or \(v = \sqrt{674}\) If inconsistent signs are used for \(u\), \(a\) and \(s\) score A0 PI AWRT \(\pm\) 25.96 If vector form is used, both components must be correct	A1	1.1b
Finds the magnitude of the resultant vector of their vertical component of velocity and 40 Do not allow 25 for their vertical component of velocity	M1	1.1a
Obtains the correct speed of the arrow. Accept AWRT 48 If inconsistent signs are used for \(u\), \(a\) and \(s\) score A0	A1	1.1b
	(4 marks)

Typical solution

\[v^2 = 25^2 + 2 - 9.8 \times -2.5\]\[v^2 = 674\]\[\text{Speed} = \sqrt{40^2 + 674}\]\[\text{Speed} = 47.7\ \text{m s}^{-1}\]\[\text{Speed} = 48\ \text{m s}^{-1}\]