Motion in a Gravitational Field: projectiles, energy and drag
Key idea: Use one uniform gravitational-field model consistently across force, component-motion and energy descriptions, then identify how drag changes the motion.
Before you start: Kinematics GraphsEnergy & Fields
By the end, you can
- Describe weight as the gravitational force on a mass and use W = mg in a uniform field.
- Explain projectile motion as uniform velocity in one direction and uniform perpendicular acceleration in the other.
- Derive ΔEₚ = mgΔh from work done and apply it with a consistent height change and sign.
- Describe falling with air resistance using resultant force, energy transfers and terminal velocity.
Starting-point self-check
1. Check your starting point
Attempt all three groups without notes and mark the first force, component or energy statement you could not justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Weight and perpendicular motion 5(a)–(b)
Question 1
Define weight. Find the weight of a 2.0 kg ball where gravitational field strength is 9.81 N kg⁻¹ and state its direction.
Check the model response
Weight is the gravitational force experienced by a mass in a gravitational field. W = mg = 2.0(9.81) = 19.6 N, directed along the gravitational field: vertically downward near Earth.
Question 2
A ball is launched horizontally at 20 m s⁻¹ from a height of 45 m. Neglect air resistance and use g = 9.8 m s⁻². Find its flight time and horizontal range.
Check the model response
Vertical motion has uᵧ = 0 and acceleration g, so 45 = ½(9.8)t² and t = 3.03 s. Horizontal velocity remains 20 m s⁻¹, so range = 20(3.03) = 60.6 m.
repair
2. Repair the five common breaks
Use only the repair matching an error, then rebuild the force, component or energy model.
Weight and perpendicular motion 5(a)–(b)
Check this idea
Misconception: Mass and weight are the same quantity.
Repair: Mass is measured in kilograms and characterises inertia. Weight is the gravitational force W = mg, measured in newtons, and changes when gravitational field strength changes.
Check this idea
Misconception: Gravity gradually reduces a projectile’s horizontal velocity when air resistance is neglected.
Repair: Gravity acts vertically, so horizontal acceleration is zero and horizontal velocity is constant. The vertical component changes with acceleration g; the shared time links the two components.
worked example
3. Follow three worked models
Follow how each solution declares axes, system boundaries and the uniform-field assumption.
Weight and perpendicular motion 5(a)–(b)
Model 1
A ball is projected horizontally at 12 m s⁻¹ from a cliff 44.1 m high. Use g = 9.8 m s⁻² and neglect air resistance. Find its time, range, impact speed and direction.
Check the model response
Take downward positive vertically. From 44.1 = ½(9.8)t², t = 3.00 s. Range = 12(3.00) = 36.0 m. At impact vₓ = 12 m s⁻¹ and vᵧ = 9.8(3.00) = 29.4 m s⁻¹ downward. Speed = √(12² + 29.4²) = 31.8 m s⁻¹ at tan⁻¹(29.4/12) = 67.8° below horizontal.
guided practice
4. Guided practice
Use each hint only to choose the force, component equation or energy-transfer direction.
Weight and perpendicular motion 5(a)–(b)
Question 1
A ball is launched at 20 m s⁻¹ and 30° above horizontal, landing at the launch height. Neglect drag and use g = 9.81 m s⁻². Find flight time and range.
Hint: Resolve the initial velocity, use vᵧ = 0 at maximum height, then double the time up.
Check the model response
uₓ = 20 cos 30° = 17.3 m s⁻¹ and uᵧ = 10.0 m s⁻¹. Flight time = 2uᵧ/g = 2.04 s. Range = uₓt = 17.3(2.04) = 35.3 m.
independent practice
5. Independent practice
Solve without repair notes and state every sign convention and modelling assumption.
Weight and perpendicular motion 5(a)–(b)
Question 1
A projectile starts with components uₓ = 18 m s⁻¹ and uᵧ = 12 m s⁻¹. After 1.5 s, find its displacement components and velocity components. Use g = 9.8 m s⁻² and neglect drag.
Check the model response
With right and up positive, x = 18(1.5) = 27 m and y = 12(1.5) − ½(9.8)(1.5²) = 6.98 m. Velocity components are vₓ = 18 m s⁻¹ and vᵧ = 12 − 9.8(1.5) = −2.70 m s⁻¹.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Weight and perpendicular motion 5(a)–(b)
Question 1
A 0.50 kg ball is launched horizontally at 14 m s⁻¹ from 19.6 m above level ground. Use g = 9.8 N kg⁻¹. Find its weight, flight time, range, impact speed and impact direction.
Check the model response
Weight = 0.50(9.8) = 4.9 N downward. From 19.6 = ½(9.8)t², t = 2.00 s and range = 14(2.00) = 28.0 m. Impact components are 14 m s⁻¹ horizontal and 19.6 m s⁻¹ downward, giving speed 24.1 m s⁻¹ at 54.5° below horizontal.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Weight and perpendicular motion 5(a)–(b)
Question 1
A 3.0 kg ball is projected at 24 m s⁻¹ and 30° above horizontal, returning to its launch height. Use g = 9.8 N kg⁻¹ and neglect drag. Find its weight, flight time and range.
Check the model response
Weight = 3.0(9.8) = 29.4 N downward. uᵧ = 12 m s⁻¹ and uₓ = 24 cos 30° = 20.8 m s⁻¹. Flight time = 2uᵧ/g = 2.45 s, so range = 20.8(2.45) = 50.9 m.