Free Pearson Edexcel International A Level Physics lesson

Projectile motion and video evidence

Two motions share one clock.

What you’ll learn

  • Apply independent horizontal and vertical motion under uniform gravity.
  • Solve horizontal and angled launches without mixing component equations.
  • Use timed images to test a projectile model and identify its limits.

Separate the axes, share the time

After a projectile leaves its launcher, ideal free motion under gravity has no horizontal force and a constant downward force mg. Choose horizontal x and upward y axes. The horizontal velocity then stays constant, while vertical velocity decreases by g each second. Both components describe one object and use the same elapsed time. A horizontal launch has zero initial vertical velocity, not zero vertical acceleration.

For a horizontal launch at speed uₓ from height h above a level landing, the vertical equation gives h = ½gt². Once that fall time is known, horizontal range is x = uₓt. Do not place horizontal speed into the vertical equation. The path curves because equal time intervals give equal horizontal steps but growing downward steps. Draw only weight on the ideal projectile’s free-body diagram after launch; a continuing forward launch force does not travel with it.

The independence is a consequence of the force model. Air resistance depends on velocity and can couple the horizontal and vertical motion; wind or lift also changes the prediction. A dropped ball and a horizontally launched ball from the same height reach a common level together only when their initial vertical velocities, release times and vertical accelerations agree. Keep these conditions explicit rather than turning a useful demonstration into a universal rule.

ax = 0; ay = −g x = uₓt y = uᵧt − ½gt²

Symbols and units

  • x and y are horizontal and upward displacements in m; uₓ and uᵧ are launch velocity components in m s⁻¹.
  • t is shared elapsed time in s; g is downward acceleration magnitude in m s⁻².

How the equation explains the idea

  1. Neglect air resistance, so horizontal acceleration is zero and vertical acceleration is −g. Integrate each constant acceleration independently to obtain x = uₓt and y = uᵧt − gt²/2.
  2. For uₓ ≠ 0, eliminate t using t = x/uₓ. Then y = (uᵧ/uₓ)x − gx²/(2uₓ²), a parabola.
  3. Changing uₓ changes horizontal range, but not fall time when the vertical launch and landing conditions are unchanged.
Horizontal speed changes range without changing fall time

Conditions. Two ideal projectiles share upward launch component 10 m s⁻¹ and return to launch height. Horizontal components are 5 and 10 m s⁻¹. Set g = 10 m s⁻² and neglect air resistance.

Horizontal displacement / m; Vertical displacement / m.

Both return after T = 2uᵧ/g = 2 s and reach 5 m height at 1 s. Doubling horizontal velocity doubles range from 10 to 20 m while leaving these vertical quantities unchanged. The curves show trajectories, not displacement against time.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

A horizontal exit from a workbench

  1. A small ball leaves a 1.25 m high bench horizontally at 3.0 m s⁻¹. Use g = 10.0 m s⁻² and ignore drag.
  2. Vertical fall time: t = √(2 × 1.25/10.0) = 0.500 s.
  3. Horizontal distance: x = 3.0 × 0.500 = 1.50 m.

Changing horizontal speed changes range but not the fall time in this ideal setup.

Keep this idea. Use the vertical motion to find the shared flight time, then use horizontal motion for the range.

Resolve an angled launch and choose the landing root

For launch speed u at angle θ above horizontal, use uₓ = u cos θ and uᵧ = u sin θ. Write a separate equation for each axis and use the vertical landing displacement to solve for time. If launch and landing heights are equal, the nonzero flight time is 2u sin θ/g. That shortcut fails when the landing is above or below the launch. A quadratic often gives more than one time; choose the root corresponding to the specified part of the flight.

At the highest point, vertical velocity is zero but horizontal velocity remains u cos θ. The projectile’s speed therefore usually is not zero at the top. At any later instant, combine vₓ and vᵧ to find speed with Pythagoras and direction with a quadrant-aware tangent ratio. Maximum height gain above the launch is uᵧ²/(2g), obtained from the vertical velocity equation without finding the horizontal position first.

With fixed launch speed, no drag and an equal-height landing, range is u² sin(2θ)/g. The greatest ideal range is at 45°, and complementary angles give the same range. These are conclusions for a particular model, not universal sports advice. Different launch and landing heights, air forces or a speed that changes with angle can move the optimum. State which assumptions justify any range comparison.

uₓ = u cos θ uᵧ = u sin θ Equal-height flight: T = 2uᵧ/g

Symbols and units

  • u is launch speed in m s⁻¹; θ is launch angle above horizontal; uₓ and uᵧ are its components.
  • T is total flight time in s and g is acceleration magnitude in m s⁻².

How the equation explains the idea

  1. Resolve uₓ = u cos θ and uᵧ = u sin θ. Set y = 0 for landing at launch height: 0 = uᵧT − gT²/2.
  2. Factor T(uᵧ − gT/2) = 0. Discard the launch root T = 0 to obtain T = 2uᵧ/g for an upward launch.
  3. The range is uₓT = u² sin(2θ)/g. This equal-height formula is not valid for a cliff landing at a different height.

Worked example

An angled launch with simple components

  1. A projectile starts with uₓ = 12 m s⁻¹ and uᵧ = 9.0 m s⁻¹. Use g = 10 m s⁻² and an equal-height landing.
  2. Time to top = 9.0/10 = 0.90 s; height gain = 9.0²/(2 × 10) = 4.05 m.
  3. Total time = 1.80 s and range = 12 × 1.80 = 21.6 m.
  4. At the top speed is 12 m s⁻¹, because the horizontal component remains.

Resolve first; a zero vertical component does not imply a stationary projectile.

Keep this idea. Use the actual landing height, and distinguish vertical velocity from the full velocity vector.

Use timed positions to test independence

A fixed camera or strobe can provide positions at known time intervals. Include a length reference in the plane of motion, keep the camera perpendicular to that plane, fix its position and identify the same point on the object in every frame. Calibrate the frame interval rather than assuming slow-motion playback time is real elapsed time. A short, safe tabletop trajectory into a catch tray is more controlled than a large launch; follow supervision and keep the flight path clear.

Tabulate horizontal and vertical coordinates with a stated origin and scale. Equal horizontal increments in equal times support constant horizontal velocity. Successive vertical increments change under gravity: with upward positive, the second difference y(t + Δt) − 2y(t) + y(t − Δt) is approximately −gΔt². Alternatively, plot y against t² for a horizontal launch, or fit y = uᵧt − ½gt² for an angled launch. The resulting table or graph is quantitative evidence, not merely a photograph that looks curved.

Finite frame spacing gives interval-average velocity; a central difference estimates instantaneous velocity at the middle time. Small position uncertainty can become a much larger relative uncertainty when differencing nearly equal coordinates, especially near the top where vertical motion is small. Compare residuals with estimated pixel/scale/time uncertainties. Horizontal slowing might indicate drag, but also check camera perspective, motion outside the scale plane and tracking errors before assigning one cause.

vₓ(midpoint) ≈ Δx/Δt aᵧ ≈ [y(t + Δt) − 2y(t) + y(t − Δt)]/Δt²

Symbols and units

  • x and y are positions in m; Δt is the time between video samples in s.
  • vₓ is horizontal velocity in m s⁻¹; aᵧ is vertical acceleration in m s⁻².

How the equation explains the idea

  1. A difference of adjacent x positions divided by Δt estimates the velocity at the interval midpoint.
  2. Subtract the velocity estimates on either side of t and divide by Δt again. This gives [y(t + Δt) − 2y(t) + y(t − Δt)]/Δt².
  3. For y = uᵧt − gt²/2 the expression equals −g exactly; noisy position measurements are amplified by division by Δt².

Worked example

Three frames reveal vertical acceleration

  1. Upward coordinates at 0.10 s intervals are y₀ = 0.000 m, y₁ = −0.050 m, y₂ = −0.200 m.
  2. Second difference = −0.200 − 2(−0.050) + 0.000 = −0.100 m.
  3. aᵧ ≈ −0.100/(0.10²) = −10.0 m s⁻². Horizontal x = 0.00, 0.30, 0.60 m gives constant 3.0 m s⁻¹.

One coordinate table can test both zero horizontal acceleration and downward vertical acceleration.

Keep this idea. Calibrate both space and time, analyse both coordinate columns and test model deviations against measurement uncertainty.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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