Projectile Motion: Complete A Level H2 Physics Cheatsheet
A Level Physics (H2), Chapter 5 · 9478 Topic 5 · Read time: ~10 minutes
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Key Formulas
Weight and free fall
Weight is the force experienced by a mass in a gravitational field, W = mg.
Without air resistance, every object near the Earth falls with the same acceleration g, whatever its mass.
Two independent motions
With air resistance neglected, a projectile has uniform velocity horizontally (no horizontal force) and uniform acceleration g vertically. The two motions are independent.
Solve them as two separate one-dimensional problems, linked only by the shared time of flight.
Method: resolve the launch velocity, find the time from the vertical motion, then horizontal range = uₓ × t.
At the highest point the vertical velocity is zero, but the horizontal velocity is still uₓ, so the speed there is not zero.
Trap: Never put a horizontal quantity into a vertical equation. Horizontally a = 0; vertically a = g. Only t links the two.
Gravitational potential energy in a uniform field
Work done lifting a mass at constant speed = force × distance = mg × Δh, so ΔEp = mgΔh.
This only holds where g is constant, near the Earth's surface. Far from the Earth, use the gravitational potential in Gravitational Fields (Chapter 8) instead.
Energy is a quick route to speeds: without air resistance, loss in Ep = gain in Ek, so ½mv² = ½mu² + mgΔh, regardless of the launch angle.
Air resistance and terminal velocity
Air resistance grows with speed. A falling object starts with acceleration g; as speed rises, drag rises, the resultant force falls and so does the acceleration.
When drag equals weight, the resultant force is zero and the object falls at terminal velocity.
In energy terms, at terminal velocity the loss of gravitational Ep goes entirely into thermal energy of the air and object, not into kinetic energy.
For a projectile with air resistance, the path is no longer symmetrical: the range and maximum height are both reduced, and the descent is steeper than the ascent.
Worked Example
A ball is kicked from ground level at 20 m s⁻¹ at 30° above the horizontal. Air resistance is negligible. Take g = 9.81 m s⁻².
(a) Calculate the time of flight. [2]
uᵧ = 20 sin 30° = 10.0 m s⁻¹
Lands when s = 0: t = 2uᵧ / g = 20.0 ÷ 9.81 = 2.04 s
(b) Calculate the horizontal range. [2]
uₓ = 20 cos 30° = 17.3 m s⁻¹
Range = 17.3 × 2.04 = 35.3 m
(c) Use energy to find the speed of the ball at its highest point, and check it against uₓ. [2]
Maximum height h = uᵧ² / 2g = 100 ÷ 19.62 = 5.10 m
½v² = ½(20)² − 9.81 × 5.10 → v² = 400 − 100 = 300, v = 17.3 m s⁻¹
This equals uₓ, as expected: only the vertical component is lost at the top.
Why this question is a good test of the topic: it solves the same flight with components and with energy. Getting the same speed at the top both ways is the check examiners like to see.
Frequently Asked Questions
Why are horizontal and vertical motion independent?
Gravity acts only vertically. With no air resistance, nothing acts horizontally, so horizontal velocity stays constant.
Is the speed zero at the top of a projectile's path?
No. Only the vertical component is zero. The horizontal component uₓ is unchanged.
When can I use ΔEp = mgΔh?
When the gravitational field is uniform, such as near the Earth's surface over small height changes.
What is terminal velocity?
The constant speed reached when air resistance equals weight, so the resultant force and acceleration are zero.
How does air resistance change a projectile's path?
It reduces both the range and maximum height, and makes the path asymmetrical, with a steeper descent.
Projectile questions taking too long?
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