Projectile motion — Question 141

Original QCE Vault practice · 9 marks

Q141 · Practice questionComplex unfamiliar9 marks

QUESTION 141 (9 marks)

A ball is kicked from level ground and lands on the flat roof of a building. Air resistance is negligible.

The graph shows the speed of the ball from the instant it is kicked until the instant it lands on the roof.

Graph of speed of the ball in metres per second against time in seconds.

Explain why the speed of the ball never falls to zero.

[2 marks]

Determine the angle above the horizontal at which the ball was kicked.

[2 marks]

Determine the height of the roof above the ground. Show your working.

[3 marks]

A second ball is kicked from the same point with the same initial speed, but at

70^\circ

above the horizontal. It also lands on the roof.

Determine the speed of the second ball when it lands.

[2 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

9 marks

ANSWER

(a) Horizontal velocity remains non-zero. (b)

53^\circ

. (c)

15\ \mathrm{m}

. (d)

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

Worked solution

Key insight. A speed–time graph of a projectile is not two straight lines: its minimum value is the horizontal component of the velocity, and the landing speed depends only on the launch speed and the height gained — not on the launch angle.

Part a · 2 marks

There is no horizontal force, so the horizontal component of velocity is constant. At the highest point the vertical component is zero, but the horizontal component is not, so the speed there is equal to the (non-zero) horizontal component.

recognises that the horizontal component of velocity is constant (no horizontal force/acceleration)

[1 mark]

identifies that at the highest point only the horizontal component remains (vertical component is zero), so speed is not zero

[1 mark]

Part b · 2 marks

From the graph: initial speed

u = 25\ \text{m s}^{-1}

; minimum speed

= u_x = 15\ \text{m s}^{-1}

(at

t \approx 2.0\ \text{s}

, the highest point).

\cos\theta = \frac{u_x}{u} = \frac{15}{25} \quad\Rightarrow\quad \theta = 53^\circ

Tolerance:

51^\circ

–

55^\circ

. Alternative:

u_y = gt_{\text{apex}} = 9.8 \times 2.04 = 20\ \text{m s}^{-1}

from the time of the minimum, then

\tan\theta = 20/15

identifies the initial speed as

u

and the minimum speed as

u_x

[1 mark]

calculates the angle

[1 mark]

Part c · 3 marks

From the graph the ball lands at

t \approx 3.09\ \text{s}

with speed

\approx 18.2\ \text{m s}^{-1}

Method 1 (components).

u_y = 25\sin 53.13^\circ = 20.0\ \text{m s}^{-1}

. At landing

v_x = 15.0

, so

v_y = -\sqrt{18.2^2 - 15.0^2} = -10.3\ \text{m s}^{-1}

(moving down, since the ball has passed its highest point).

v_y^2 = u_y^2 + 2 a s_y \;\Rightarrow\; s_y = \frac{u_y^2 - v_y^2}{2g} = \frac{20.0^2 - 10.3^2}{2 \times 9.8} = 15\ \text{m}

Method 2 (time).

s_y = u_y t + \tfrac12 a t^2 = 20.0(3.09) - 4.9(3.09)^2 = 15\ \text{m}

Tolerance:

14

–

16\ \text{m}

(reading the end of the curve to

\pm 0.2\ \text{m s}^{-1}

gives

14.6

–

15.4\ \text{m}

; reading

t

\pm 0.05\ \text{s}

gives

14.4

–

15.4\ \text{m}

). Allow FT from b).

recognises the appropriate relationship for vertical motion (or

v^2 = u^2 - 2gh

)

[1 mark]

determines the required quantities from the graph and the earlier part (final speed or time;

u_y

u_x

)

[1 mark]

calculates the height

[1 mark]

Part d · 2 marks

u_x = 25\cos 70^\circ = 8.55\ \text{m s}^{-1}

u_y = 25\sin 70^\circ = 23.49\ \text{m s}^{-1}

. (This ball rises to

28.2\ \text{m}

, above the roof, and lands on the way down.)

v_y^2 = u_y^2 - 2gh = 23.49^2 - 2(9.8)(15) = 257.9

v = \sqrt{u_x^2 + v_y^2} = \sqrt{8.55^2 + 257.9} = 18.2\ \text{m s}^{-1}

The same landing speed as the first ball. Because

v^2 = v_x^2 + v_y^2 = (u_x^2 + u_y^2) - 2gh = u^2 - 2gh

, the landing speed depends only on the launch speed and the height of the roof.

Tolerance:

18.2 \pm 0.3\ \text{m s}^{-1}

; if the roof height from c) is

14

–

16\ \text{m}

then

17.6

–

18.7\ \text{m s}^{-1}

is acceptable (FT).

calculates the vertical component of velocity at the roof using the roof height

[1 mark]

combines components to calculate the speed

[1 mark]

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

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Projectile motion — Question 141

QUESTION 141 (9 marks)

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