EXN.2.SL.TZ0.3

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The rate of change of the height $(h)$ of a ball above horizontal ground, measured in metres, $t$ seconds after it has been thrown and until it hits the ground, can be modelled by the equation

$\frac{d h}{d t} = 11.4 - 9.8 t$

The height of the ball when $t = 0$ is $1.2 m$ .

Find an expression for the height $h$ of the ball at time $t$ .

[6]

Find the value of $t$ at which the ball hits the ground.

[2]

b.i.

Hence write down the domain of $h$ .

[1]

b.ii.

Find the range of $h$ .

[3]

Markscheme / solution

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

$h = \int (11.4 - 9.8 t) d t$ M1

$h = 11.4 t - 4.9 t^{2} (+ c)$ A1A1

When $t = 0, h = 1.2$ (M1)

$c = 1.2$ (A1)

$(h =) 1.2 + 11.4 t - 4.9 t^{2}$ A1

[6 marks]

$2.43 (2.42741 \dots)$ seconds (M1)A1

[2 marks]

b.i.

$0 \leq t \leq 2.43$ A1

Note: Accept $0 \leq t < 2.43$ .

[1 mark]

b.ii.

Maximum value is $7.83061 \dots$ (M1)

Range is $0 \leq h \leq 7.83$ A1A1

Note: Accept $0 \leq h < 7.83$ .

[3 marks]

Examiners’ report

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b.i.

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b.ii.

[N/A]