Higher June 2018 Unit 1 Q13
13. A ball is thrown upwards from a height of 1 m above the ground.
After \(t\) seconds, its height above the ground is \(h\) metres, where \(h\) is given by
\[h = 1 + 8t - 5t^2.\](a) Show that the time taken to reach a height of 4 metres satisfies the equation\[5t^2 - 8t + 3 = 0.\]
[1]
(b) Solve the equation\[5t^2 - 8t + 3 = 0.\]
[3]
(c) Interpret your answer to part (b) in the context of the question. [1]
From the formula list: The solutions of \(ax^2 + bx + c = 0\) where \(a \neq 0\) are given by \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
| Scheme | Marks | Comments |
|---|---|---|
| \(4 = 1 + 8t - 5t^2\) or \(1 + 8t - 5t^2 = 4\) leading to \(5t^2 - 8t + 3 = 0\) | B1 | Must be convincing. |
| Scheme | Marks | Comments |
|---|---|---|
| \((5t - 3)(t - 1)\ (=0)\) | B2 | B1 for \((5t \;….\; 3)(t \;…..\; 1)\) |
| (t =) 3/5 AND 1 | B1 | Strict FT from ‘their two brackets’. (Both solutions are required for this B1.) |
Using quadratic formula.
| Scheme | Marks | Comments |
|---|---|---|
| \((t =)\ \dfrac{8 \pm \sqrt{[(-8)^2 - 4(5)(3)]}}{2(5)}\) | M1 | Allow one error, in sign or substitution, but not in the formula. |
| \(t = \dfrac{8 \pm \sqrt{4}}{10}\) | A1 | |
| t = 3/5 AND 1 | A1 |
Using trial and improvement
| Scheme | Marks | Comments |
|---|---|---|
| Award B3 for a method leading to both solutions, namely t = 3/5 AND t = 1, otherwise B0. |
| Scheme | Marks | Comments |
|---|---|---|
| Valid statement | E1 | e.g. 2 different values of t representing the ball on its way up and on its way down OR e.g. the ball reaches its highest point after 4/5 s. FT provided both solutions are positive. |