Gravity and motion — Question 155

Original QCE Vault practice · 8 marks

Q155 · Practice questionComplex unfamiliar8 marks

QUESTION 155 (8 marks)

A ball leaves a platform

5.00\,\mathrm m

above level ground with velocity components

u_x=16.0\,\mathrm{m\,s^{-1}}

and

u_y=12.0\,\mathrm{m\,s^{-1}}

. A thin wall is

24.0\,\mathrm m

horizontally from the launch point and

11.0\,\mathrm m

high. Ignore drag and the ball's size. Use

g=9.8\,\mathrm{m\,s^{-2}}

A ball launches from the horizontal centre of a platform 5.00 m high at the left. Launch velocity points up and right. Dotted vertical guides align the launch point and near face of an 11.0 m wall with the 24.0 m horizontal dimension below the ground line. Not to scale.

Determine whether the ball clears the wall. Justify quantitatively.

[3 marks]

Determine the speed and direction of the ball when it passes the wall.

[3 marks]

Calculate the horizontal distance from the launch point to the ground impact.

[2 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

8 marks

ANSWER

(a) It clears by

0.975\,\mathrm m

(about

0.98\,\mathrm m

). (b)

16.2\,\mathrm{m\,s^{-1}}

9.58^\circ

below horizontal. (c)

45.0\,\mathrm m

Worked solution

(a)

t=x/u_x=24/16=1.50\,\mathrm s

. Height above ground is

y=5+12(1.50)-4.9(1.50)^2=11.975\,\mathrm m

. Thus clearance is

11.975-11.0=0.975\,\mathrm m>0

(b)

v_x=16.0

and

v_y=12.0-9.8(1.50)=-2.70\,\mathrm{m\,s^{-1}}

. Hence

v=\sqrt{16^2+2.7^2}=16.226\,\mathrm{m\,s^{-1}}

and

\alpha=\tan^{-1}(2.7/16)=9.578^\circ

downward.

0=5+12t-4.9t^2

. The positive root is

t=2.81187\,\mathrm s

, so

x=16t=44.98996\,\mathrm m

. The negative root is before launch.

Equivalent physically justified methods and consistent equivalent units accepted.

Displayed decimals are model answers; accept appropriate significant figures and consistent rounding from stated constants.

Part a: Determine the time at the wall.

[1 mark]

Part a: Calculate height above ground with the launch height included.

[1 mark]

Part a: Compare with wall height and conclude it clears.

[1 mark]

Part b: Calculate the vertical component with its sign.

[1 mark]

Part b: Combine the perpendicular components for speed.

[1 mark]

Part b: Determine an angle below horizontal.

[1 mark]

Part c: Select the positive ground-impact time.

[1 mark]

Part c: Calculate the horizontal distance.

[1 mark]

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

View the QCAA syllabus

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Gravity and motion — Question 155

QUESTION 155 (8 marks)

Practice marking scheme

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