June 2022 Paper 2 Q13
13 In this question use \(g\) = 9.8 m s−2
A ball is projected from a point on horizontal ground with an initial velocity of 7 m s−1 at an angle \(\theta\) above the horizontal.
The ball reaches a maximum vertical height of \(h\) metres above the ground.
(a) Show that\[h = 2.5\sin^2\theta\] [3 marks]
(b) Hence, given that \(0^\circ \leqslant \theta \leqslant 60^\circ\), find the maximum value of \(h\). [2 marks]
(c) Nisha claims that the larger the size of the ball, the greater the maximum vertical height will be.
State whether Nisha is correct, giving a reason for your answer. [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| Obtains \(7\sin\theta\) for vertical component of initial velocity | B1 | 1.1b |
| Uses \(v^2 = u^2 + 2as\) with \(v = 0\) Or uses appropriate constant acceleration equations that form a complete method to obtain \(h\) eg finds \(t\) then subs into \(s = ut + \dfrac{1}{2}at^2\) | M1 | 3.3 |
| Completes argument substituting \(s = h\), \(v = 0\), \(u = 7\sin\theta\) and \(a = -9.8\) to show the given result. Accept negative values used consistently AG | R1 | 2.1 |
| (3) |
Typical solution
\[u = 7\sin\theta\]\[v^2 = u^2 + 2as\]\[0 = 49\sin^2\theta - 19.6h\]So
\[h = 2.5\sin^2\theta\]| Scheme | Marks | AO |
|---|---|---|
| Substitutes any value of \(\theta\) in the range \(0 \lt \theta \leqslant 60\) to obtain a height greater than zero | M1 | 1.1a |
| Deduces \(h\) = 1.9 AWRT 1.9 or \(\dfrac{15}{8}\) | A1 | 2.2a |
| (2) |
Typical solution
\[\theta = 60^\circ\]\[h = 2.5\sin^2 60\]\[h = 1.9\]| Scheme | Marks | AO |
|---|---|---|
| States that Nisha is incorrect and refers to air resistance or ball modelled as a particle | E1 | 3.5a |
| (1) | ||
| (6 marks) |
Typical solution
Nisha is incorrect
The model ignores air resistance which gets greater as ball gets larger