A-level · Beta

Projectile motion

Change a launch and connect its path with horizontal and vertical velocity.

Change the launch, then move the time slider to inspect the projectile and velocity components.

x=ucos⁡(θ)t,y=h+usin⁡(θ)t−12gt2x=u\cos(\theta)t,\quad y=h+u\sin(\theta)t-\tfrac12gt^2
Change the launch, then move the time slider to inspect the projectile and velocity components.0011.7353.061223.4696.122435.2049.183746.93912.245Horizontal distance (m)Height (m)

At 1.1545 s: position (16.327, 9.7959) m.

Time (s)
1.1545
Horizontal velocity (m/s)
14.142
Vertical velocity (m/s)
2.8284
Range (m)
40.816
Greatest height (m)
10.204
Flight time (s)
2.8862

Constant gravity g = 9.8 m/s²; no air resistance. Green lines show velocity components at a separate drawing scale. Axes may have different scales. The maximum-range 45° rule requires equal launch and landing heights.

Predict, test and explain

Step 1 of 3

Predict

At the highest point of a 45° launch, is the whole velocity zero?

InvestigateCompare complementary angles for a launch from ground level. What stays equal?

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Try it in the simulation. Change one thing at a time.

Notes and saved values stay in this tab. They disappear when you reload.

Hints, self-check and connections
Try a self-check

Fixed values, separate from the diagram controls.

At the highest point of a 45° launch, is the whole velocity zero?

vy=0v_y=0

Hint

Separate horizontal and vertical equations; time is shared.

Reveal answer

Gravity changes the vertical component. The horizontal component stays constant in this model.

vx=ucos⁡45∘≠0v_x=u\cos45^\circ\ne0

Build an explanation

Hint 1

Separate horizontal and vertical equations; time is shared.

Hint 2

Gravity changes the vertical component. The horizontal component stays constant in this model.

Hint 3

Use the model to check a prediction, then explain the result.

Worked example

This example uses fixed values, separate from the diagram controls.

A projectile is launched at 14 m/s at 45° from ground level. Find its range when g = 9.8 m/s².

  1. Resolve the initial velocity into 14cos45° and 14sin45°.
  2. Use the vertical equation to find the positive landing time.
  3. Substitute that time into the horizontal equation, giving u²sin(2θ)/g.

Answer: 20 m.

Connect this idea

Watch out: Gravity changes the vertical component. The horizontal component stays constant in this model.

Specification and learning route

AQA Q5 · Edexcel Mechanics 7.5

A-level extension. This model illustrates selected content; references are not a claim of full coverage or exam-board endorsement.

Useful starting knowledge: Straight-line suvat equations; Vector components; Trigonometry; Acceleration under gravity.

AQA specification · Edexcel specification

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