Projectile motion
Resolve H1 Physics projectile motion into horizontal and vertical components, calculate gravitational energy changes and explain air resistance.
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Before you begin
Near the Earth’s surface the gravitational field is uniform, so the weight of an object is constant. The lessons use this to analyse projectiles by treating their horizontal and vertical motions separately, then derive the change in gravitational potential energy from the work done by gravity, and finish with how air resistance leads to terminal velocity.
Be comfortable with: uniformly accelerated motion, because the vertical motion of a projectile is uniformly accelerated, and kinetic energy.
Learning goals
- Relate weight and gravitational potential energy changes in a uniform gravitational field.
- Analyse projectile motion by separating perpendicular components.
- Explain falling motion with air resistance using forces, energy and terminal velocity.
Syllabus statements covered
- describe and use the concept of weight as the force experienced by a mass in a gravitational field
- describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction
- derive, from the definition of work done by a force, the equation ∆Ep = mg∆h for gravitational potential energy changes in a uniform gravitational field (e.g. near the Earth’s surface)
- recall and use the equation ∆Ep = mg∆h to solve problems
- describe qualitatively, with reference to forces and energy, the motion of bodies falling in a uniform gravitational field with air resistance, including the phenomenon of terminal velocity
Lessons
Work through them in order.
- Projectile motion in perpendicular componentsUse weight as the force in a gravitational field, and solve projectile problems by treating the horizontal and vertical motions separately.
- Potential energy in a uniform gravitational fieldDerive ΔEp = mgΔh from the work done by gravity, and use it, explaining why the route between two heights does not matter.
- Air resistance and terminal velocityDescribe, with forces and energy, how an object falling through air approaches its terminal velocity.
Practise and check
Or choose
Topic reference
Quick reference
With up as positive and no air resistance:
| Quantity | Horizontal | Vertical |
|---|---|---|
| initial velocity | uₓ = u cos θ | uy = u sin θ |
| acceleration | 0 | -g |
| velocity at time t | vₓ = uₓ | vy = uy-gt |
| displacement at time t | x = uₓt | y = uyt-(1/2)gt² |
| Idea | Relationship |
|---|---|
| weight | W = mg, with g = 9.81 N kg⁻¹ |
| gravitational potential energy | Δ Eₚ = mgΔ h, Δ h = h_final-hᵢₙᵢₜᵢₐₗ |
| falling with drag (down positive) | ma = mg-D |
| terminal velocity | D = mg, so a = 0 |
Core knowledge to remember
- Weight is the force on a mass in a gravitational field, W = mg.
- With no air resistance, a projectile has constant horizontal velocity and a constant downward acceleration g. The two motions are independent but last the same time, and the path is a parabola.
- In a uniform field, the change in gravitational potential energy is Δ Eₚ = mgΔ h, derived from the work done by gravity. It depends only on the change in height, not on the route.
- An object falling through air has a drag force that increases with speed. Its acceleration decreases until the drag equals its weight; it then falls at its terminal velocity.
- At terminal velocity the kinetic energy is constant, but the object keeps losing gravitational potential energy, which is transferred by drag to internal energy of the air and surroundings.
Common mistakes
- Thinking gravity reduces the horizontal velocity of a projectile.
- Saying the acceleration is zero at the highest point of a flight.
- Using different times for the horizontal and vertical motions.
- Using the distance along a path instead of the vertical height change in mgΔ h.
- Saying no forces act at terminal velocity, or that energy transfer stops.