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How do you analyse projectile motion by treating the horizontal and vertical components separately?

Model projectile motion: resolve initial velocity into horizontal and vertical components, treat horizontal motion at constant velocity and vertical motion under gravity, and find range, time of flight and maximum height.

A CCEA GCSE Further Mathematics answer on projectile motion, covering resolving the launch velocity into components, constant horizontal velocity, vertical motion under gravity, and finding the time of flight, range and maximum height in the Mechanics unit.

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  1. What this dot point is asking
  2. Resolving the launch velocity
  3. Horizontal motion
  4. Vertical motion
  5. Key features of the flight
  6. Why this matters

What this dot point is asking

A projectile is an object thrown or launched that then moves freely under gravity, and CCEA GCSE Further Mathematics analyses its curved path by splitting it into two simple motions. You must resolve the initial velocity into horizontal and vertical components, treat the horizontal motion as constant velocity and the vertical motion as constant acceleration under gravity, and find the time of flight, the horizontal range and the maximum height. The independence of the two directions is the central idea.

Resolving the launch velocity

The launch velocity is a single vector at an angle, but the two directions behave differently, so the first step is always to split it into components using trigonometry.

For a projectile launched horizontally, the angle is zero, so the whole speed is horizontal and the initial vertical velocity is zero, which simplifies the vertical equations.

Horizontal motion

Once airborne, nothing accelerates the projectile horizontally (air resistance is ignored in this model), so the horizontal velocity is unchanged for the whole flight. The horizontal distance is therefore simply the horizontal velocity multiplied by the time.

Vertical motion

Vertically the projectile is in free fall, with acceleration gg directed downward, so the suvat equations apply with a=ga = -g when up is taken as positive. The vertical velocity decreases on the way up, is zero at the highest point, and increases downward on the way down.

Key features of the flight

Three quantities are asked for repeatedly. The maximum height is found from the vertical motion by setting the vertical velocity to zero, since the projectile is momentarily moving horizontally at the top. The time of flight is the time for the vertical displacement to return to its launch level. The range is the horizontal velocity multiplied by the time of flight. Every one of these uses the shared time as the link between the two directions, which is why finding the time is so often the pivotal step.

Why this matters

Projectile motion is the showcase application of the Mechanics unit, combining the resolving of vectors, the suvat equations and trigonometry into one model. The independence of horizontal and vertical motion is a powerful idea that recurs throughout physics and applied mathematics. Mastering the routine of resolve, find the time vertically, then apply it horizontally gives a reliable method for the full range of projectile questions.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA Unit 2 (style)4 marksA ball is projected horizontally at 1414 m/s from a height of 2020 m. Taking g=9.8g = 9.8 m/s2^2, find the time to reach the ground and the horizontal distance travelled.
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Vertical motion: the ball starts with zero vertical velocity, so u=0u = 0, a=9.8a = 9.8, s=20s = 20. Use s=ut+12at2s = ut + \tfrac{1}{2}at^2: 20=0+4.9t220 = 0 + 4.9t^2, so t2=204.9=4.08t^2 = \dfrac{20}{4.9} = 4.08 and t=2.02t = 2.02 s.

Horizontal motion: constant velocity 1414 m/s, so distance =14×2.02=28.3= 14 \times 2.02 = 28.3 m.

Marks are for the vertical equation, the time, and the horizontal distance. The horizontal and vertical motions are independent and share only the time.

CCEA Unit 2 (style)6 marksA particle is projected at 2525 m/s at 3030^\circ above the horizontal. Taking g=9.8g = 9.8 m/s2^2, find the maximum height reached and the time of flight.
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Resolve the launch velocity: horizontal 25cos30=21.6525\cos 30^\circ = 21.65 m/s; vertical 25sin30=12.525\sin 30^\circ = 12.5 m/s.

Maximum height: at the top the vertical velocity is zero. Use v2=u2+2asv^2 = u^2 + 2as with v=0v = 0, u=12.5u = 12.5, a=9.8a = -9.8: 0=156.2519.6s0 = 156.25 - 19.6s, so s=7.97s = 7.97 m.

Time of flight: vertical displacement returns to zero. Use s=ut+12at2s = ut + \tfrac{1}{2}at^2 with s=0s = 0: 0=12.5t4.9t2=t(12.54.9t)0 = 12.5t - 4.9t^2 = t(12.5 - 4.9t), so t=0t = 0 or t=2.55t = 2.55 s. The time of flight is 2.552.55 s.

Marks are for resolving, the maximum height, and the time of flight.

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