4MA0 Higher January 2016 Paper 3R Q12
12 A stone is thrown vertically upwards from a point \(O\).
The height above \(O\) of the stone \(t\) seconds after it was thrown from \(O\) is \(h\) metres, where \(h = 17t - 5t^2\)
Work out the values of \(t\) when the height of the stone above \(O\) is 12 metres.
Show your working clearly.
(3)
| Scheme | Marks |
|---|---|
| \(17t - 5t^2 = 12\) \(5t^2 - 17t + 12\;(= 0)\) | M1 |
Eg \((5t - 12)(t - 1)\;(= 0)\) or \((t =)\;\dfrac{--17 \pm \sqrt{(-17)^2 - 4 \times 5 \times 12}}{2 \times 5}\) or \(\dfrac{17 \pm \sqrt{49}}{10}\) or \(\dfrac{17 \pm 7}{10}\) | M1 |
| \(t = 1\) \(t = 2.4\) oe | A1 |
| (3) |
Notes
M1: Correct three term quadratic.
e.g. \(17t - 5t^2 = 12\)
M1: For \((5t \pm 12)(t \pm 1)\) or for a correct substitution into the quadratic formula (condone 1 sign error)
Condone \(17^2\) and omission of brackets around \(-17\)
A1: Dep on M2