Solved: The rate of change of the height (h) of a ball above horizontal ground, measured... [IB Mathematics Applications & Interpretation (Math AI)]

Question

SLPaper 2

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 theequation

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

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]

Verified

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]

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

[2]

Verified

Solution

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

[2 marks]

Hence write down the domain of $h$ .

[1]

Verified

Solution

$0 \leq t \leq 2.43$ A1

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

[1 mark]

Find the range of $h$ .

[3]

Verified

Solution

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]

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 theequationdhdt=11.4-9.8tThe height of the ball whent=0is1.2 m.

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