Modelling motion — Question 225

Original QCE Vault practice · 10 marks

Q225 · Practice questionTechnology-activeComplex unfamiliar10 marks

QUESTION 225 (10 marks)

An object is thrown upwards with v(0)=10 m/s. Upwards is positive. Linear air resistance gives

dv/dt=-10-v

throughout flight; h(0)=0.

Height curve of a thrown object under linear drag, from launch at ground to its first return to ground.

Determine v(t) and h(t).

[3 marks]

Determine maximum height exactly.

[2 marks]

Find the first positive return-to-ground time to four decimal places.

[2 marks]

A tracking camera measures mean height during the complete flight:

\overline h=T^{-1}\int_0^Th(t)\,dt

, where T is the return time. Derive

\overline h=10-5T

and determine whether it exceeds 2 m.

[3 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

10 marks

ANSWER

(a)

v=20e^{-t}-10

;

h=20(1-e^{-t})-10t

. (b)

10(1-\ln2)

m. (c)

1.5936

s. (d)

\overline h=10-5T\approx2.0319

m, so it exceeds 2 m.

Worked solution

(a) Separation of

v'=-10-v

gives

v+10=Ce^{-t}

. Initial velocity gives C=20. Integrating and using zero initial height gives the stated h.

(b) At the apex v=0, so t=ln2. Substitution gives

h=20(1-1/2)-10\ln2=10(1-\ln2)

. Velocity changes from positive to negative.

2(1-e^{-t})-t=0

numerically and exclude the initial root t=0. The positive root is 1.5936. Height decreases strictly after ln2, so this is the first return.

(d) Integrating

h(t)=20(1-e^{-t})-10t

gives

\int_0^Th(t)\,dt=20T-20(1-e^{-T})-5T^2

. At return,

h(T)=0

implies

20(1-e^{-T})=10T

. Hence the integral simplifies to

10T-5T^2

and division by T gives

10-5T

. Using the unrounded root T approximately 1.593624260 gives mean height approximately 2.0319 m.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Solve velocity with the initial condition.

[1 mark]

Part a: Integrate velocity.

[1 mark]

Part a: Apply initial height.

[1 mark]

Part b: Find apex time.

[1 mark]

Part b: Evaluate the maximum height.

[1 mark]

Part c: Set the ground equation and exclude zero.

[1 mark]

Part c: Find and justify the first positive root.

[1 mark]

Part d: Integrate height over the full flight.

[1 mark]

Part d: Use the return condition to obtain the required expression.

[1 mark]

Part d: Evaluate and assess the 2 m threshold.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

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Modelling motion — Question 225

QUESTION 225 (10 marks)

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