· H2 Physics — Ten-Year Series, worked ← All 8 questions

2023 A-Level · Paper 2 · Question 2

The slowest point of a projectile's flight, and what it gives away

Kinematics · projectile motion 7 marks Kinetic energy Air resistance

An object is launched from level ground at 42° above the horizontal. The only speed you are given is hidden in an energy: the smallest kinetic energy it has anywhere on its path. Seven marks: the vertical component of the launch velocity, the range, and what air resistance does to the times going up and coming down. Below is the full solution, and beside it the flight itself, with and without air.

The model, live

Horizontal componentVertical componentPath / kinetic energyWith air resistance

Part (a)

3 marks

The object has mass 0.55 kg0.55\,\text{kg} and its minimum kinetic energy on the path is 65 J65\,\text{J}. Show that the vertical component of the launch velocity is 14 m s−114\,\text{m s}^{-1}.

Approach

With no air resistance the horizontal component of velocity never changes, and the vertical component passes through zero at the top. So the speed, and the kinetic energy, are smallest at the highest point, where the velocity is purely horizontal. The minimum kinetic energy therefore gives the horizontal component directly; the launch angle converts it to the vertical one.

At the highest point the velocity is horizontal and equal to vxv_x:

12(0.55) vx2=65⟹vx=2(65)0.55=15.37 m s−1\tfrac12 (0.55)\,v_x^{2} = 65 \quad\Longrightarrow\quad v_x = \sqrt{\frac{2(65)}{0.55}} = 15.37\,\text{m s}^{-1}

The launch velocity makes 42∘42^\circ with the horizontal, so

vy=vxtan⁡42∘=(15.37)(0.9004)=13.8 m s−1v_y = v_x \tan 42^\circ = (15.37)(0.9004) = 13.8\,\text{m s}^{-1}

To check your answer:

The launch speed is 15.372+13.82=20.7 m s−1\sqrt{15.37^2 + 13.8^2} = 20.7\,\text{m s}^{-1}, so the launch kinetic energy is 12(0.55)(20.7)2=117 J\tfrac12(0.55)(20.7)^2 = 117\,\text{J}. All of the drop from 117 J117\,\text{J} to 65 J65\,\text{J} is the vertical part, 12(0.55)(13.8)2=52 J\tfrac12(0.55)(13.8)^2 = 52\,\text{J}, which becomes gravitational potential energy at the top.

Kinetic energy is least at the top, where v=vxv = v_x: 12(0.55)vx2=65\tfrac12(0.55)v_x^2 = 65 gives vx=15.4 m s−1v_x = 15.4\,\text{m s}^{-1}. Then vy=vxtan⁡42∘=13.8 m s−1≈14 m s−1v_y = v_x\tan 42^\circ = 13.8\,\text{m s}^{-1} \approx 14\,\text{m s}^{-1}.

Watch

The minimum kinetic energy is not zero. Only the vertical component vanishes at the top; the object is still moving sideways at 15.4 m s−115.4\,\text{m s}^{-1}.

Part (b)

3 marks

Find the horizontal displacement from the launch point to where the object lands.

Approach

Treat the two directions separately. The vertical motion alone sets the time of flight; the horizontal motion is uniform, so the range is vxv_x times that time.

Vertically, taking upwards as positive, the object returns to its launch level when sy=0s_y = 0:

0=vyt−12gt2⟹t=2vyg=2(13.84)9.81=2.82 s0 = v_y t - \tfrac12 g t^2 \quad\Longrightarrow\quad t = \frac{2v_y}{g} = \frac{2(13.84)}{9.81} = 2.82\,\text{s}

Horizontally, at constant velocity:

sx=vxt=(15.37)(2.82)=43.4 ms_x = v_x t = (15.37)(2.82) = 43.4\,\text{m}

To check your answer:

The range formula R=v2sin⁡2θg=(20.7)2sin⁡84∘9.81=43 mR = \dfrac{v^2\sin 2\theta}{g} = \dfrac{(20.7)^2 \sin 84^\circ}{9.81} = 43\,\text{m} agrees.

displacement =43 m= 43\,\text{m}

Note

If you carry the rounded 14 m s−114\,\text{m s}^{-1} forward instead of 13.813.8, the time becomes 2.85 s2.85\,\text{s} and the range 44 m44\,\text{m}. Keep the unrounded value from (a) for the later working.

Part (c)

1 mark

With air resistance, compare the time taken to reach the maximum height with the time taken to fall back to the ground, giving a reason.

Approach

Compare the vertical accelerations in the two halves. Air resistance always opposes the velocity, so its vertical part points down while the object rises and up while it falls.

Going up, the weight and the vertical part of the air resistance both act downwards, so the object slows at more than gg. Coming down, the air resistance acts upwards against the weight, so it speeds up at less than gg. It covers the same height both ways, but slowly on the way down, so the descent takes longer.

The time to reach the maximum height is shorter than the time to fall back. On the way up, air resistance acts downwards with the weight, so the deceleration is greater than gg; on the way down, air resistance acts upwards, so the acceleration is less than gg.

Note

The same reasoning, in energy terms: work is done against air resistance throughout the flight, so at every height the object is moving more slowly on the way down than it was on the way up.

What this question was really testing

  1. Where the minimum is. With no air resistance, the minimum speed is at the top and equals the horizontal component. The energy data unlocks the whole question.
  2. Components. vy=vxtan⁡θv_y = v_x\tan\theta. Resolving correctly is worth more here than any formula.
  3. Independent motions. The vertical motion fixes the time; the horizontal motion uses it.
  4. The direction of air resistance. It opposes velocity, so it helps gravity going up and fights it coming down.

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