Free Fall and Projectile Motion

Spec mapping: OCR H556 Module 3.1 — Motion (motion under gravity, projectile motion, horizontal and vertical components of velocity, independence of horizontal and vertical motion). PAG 6 anchor: free-fall determination of $g$ . Refer to the official OCR H556 specification document for exact wording.

Gravity is the simplest force in nature to model and yet generates an outsized share of A-Level mechanics problems. Near the Earth's surface, any body in free fall experiences a constant downward acceleration of magnitude $g \approx 9.81$ m s $^{-2}$ , regardless of its mass. This is the cornerstone of Galileo's famous result — paraphrased rather than quoted — that, in the absence of air resistance, a feather and a cannonball fall side by side.

This lesson extends the SUVAT toolkit of the previous lesson to vertical free fall and then to two-dimensional projectile motion. The central physical insight, due to Galileo, is that horizontal and vertical motion are independent: they share the same clock but nothing else. Every projectile problem therefore reduces to two coupled one-dimensional SUVAT problems linked only by a common time.

Key Definition: A body is in free fall when the only force acting on it is its weight $W = mg$ . The acceleration of any such body equals the local gravitational field strength $g$ — independent of mass, by virtue of Newton's second law: $a = F/m = mg/m = g$ .

Free Fall — Definitions and Universality

A body is in free fall when:

Air resistance is negligible (no drag).
Buoyancy is negligible (rare on Earth; relevant only for very low-density objects in atmosphere).
No applied force acts other than gravity.

The acceleration under free fall is the gravitational field strength $g$ , measured in m s $^{-2}$ or equivalently N kg $^{-1}$ . The two units are identical: $g = F_{\text{gravity}}/m$ has units N kg $^{-1}$ ; $g =$ acceleration has units m s $^{-2}$ ; and since 1 N = 1 kg m s $^{-2}$ , the two are dimensionally the same. This dual character — gravitational field strength and free-fall acceleration — is a consequence of Newton's second law applied to a body whose only force is its own weight.

Location	$g$ (m s $^{-2}$ )
UK (typical)	9.81
Equator (sea level)	9.78
Poles (sea level)	9.83
Top of Mt Everest	9.77
ISS orbit (400 km altitude)	8.69
Moon (surface)	1.62
Mars (surface)	3.71
Jupiter (cloud tops)	24.8
Sun (photosphere)	274

Exam Tip: OCR's data sheet quotes $g = 9.81$ m s $^{-2}$ unless an alternate value is explicitly given in the question. Do not default to 10 m s $^{-2}$ for "easier arithmetic" — the marking tolerance is too narrow.

Why mass cancels — the universality of free fall

Newton's second law: $F = ma$ . For a freely falling body, the only force is gravity, $F = mg$ . So:

$ma = mg \quad \Longrightarrow \quad a = g$

The mass $m$ cancels. Every object experiences the same acceleration regardless of mass — provided air resistance is negligible. This famous result is sometimes attributed to Galileo's apocryphal Leaning-Tower-of-Pisa experiment (more likely a thought experiment than an actual one). It was beautifully verified on the airless Moon in 1971 by Apollo 15 astronaut David Scott, who dropped a hammer and a feather side-by-side; both struck the lunar surface simultaneously, captured on film for a public broadcast.

Modern precision tests — the so-called Eötvös experiments and their 21st-century descendants such as the satellite-based MICROSCOPE mission — have confirmed the universality of free fall (the weak equivalence principle) to a fractional precision better than $10^{-15}$ . Any deviation from universality would imply new physics beyond Einstein's general relativity.

Vertical Free Fall — Worked Examples

Worked Example 1 — Dropped Stone

A stone is dropped from a cliff (initial velocity zero). Ignoring air resistance, find (a) its speed after $2.5$ s and (b) the distance fallen.

Take downward positive, so $u = 0$ , $a = +9.81$ m s $^{-2}$ , $t = 2.5$ s.

(a) $v = u + at = 0 + 9.81 \times 2.5 = 24.5$ m s $^{-1}$ .

(b) $s = ut + \tfrac{1}{2} a t^2 = 0 + \tfrac{1}{2} \times 9.81 \times 6.25 = 30.7$ m.

Sanity check via $v^2 = u^2 + 2as$ : $24.5^2 = 600$ , $2 \times 9.81 \times 30.7 = 602$ — agreement to better than 1% (the tiny gap is rounding).

Worked Example 2 — Upward Throw and Symmetry

A ball is thrown vertically upward at $u = 20$ m s $^{-1}$ from ground level. Take upward positive: $a = -9.81$ m s $^{-2}$ .

Time to maximum height ( $v = 0$ ):

$0 = 20 + (-9.81) t \Rightarrow t = 2.04 \text{ s}$

Maximum height ( $v = 0$ ):

$0 = 20^2 + 2(-9.81)\,s \Rightarrow s = \frac{400}{19.62} = 20.4 \text{ m}$

Total time of flight (returns to $s = 0$ ):

$0 = 20\,t + \tfrac{1}{2}(-9.81)\,t^2 = t(20 - 4.905\,t) \Rightarrow t = 4.08 \text{ s}$

Note that the time of ascent (2.04 s) equals the time of descent (4.08 - 2.04 = 2.04 s). This symmetry of free fall is a powerful shortcut: in the absence of air resistance, time up equals time down, and the speed at any height during ascent equals the speed at that same height during descent. The graph is a triangle on a $v$ – $t$ plot, with the apex at $(2.04, 0)$ .

Worked Example 3 — Asymmetric drop from a cliff

A ball is thrown vertically upward at $u = 15$ m s $^{-1}$ from a cliff $20$ m above the sea below. Find the time of flight until it hits the water.

Take upward positive; ball lands at $s = -20$ m.

$-20 = 15\,t - 4.905\,t^2 \Rightarrow 4.905\,t^2 - 15\,t - 20 = 0$

$t = \frac{15 \pm \sqrt{225 + 392.4}}{9.81} = \frac{15 \pm 24.85}{9.81}$

Physical root: $t = 39.85 / 9.81 = 4.06$ s. The negative root corresponds to the (unphysical) extrapolation backward in time.

The ball reaches the sea $4.06$ s after launch. Speed at impact: $v = u + at = 15 - 9.81 \times 4.06 = -24.85$ m s $^{-1}$ , i.e. $24.9$ m s $^{-1}$ downward.

PAG 6 — Measuring $g$ Experimentally

OCR's Practical Activity Group 6 (Newtonian mechanics) includes the free-fall determination of $g$ . The standard method uses an electromagnet to release a steel ball, with a trapdoor switch or pair of light gates timing the fall through a known distance $h$ .

From $s = ut + \tfrac{1}{2} a t^2$ with $u = 0$ and $a = g$ :

$h = \tfrac{1}{2} g t^2 \quad \Longrightarrow \quad g = \frac{2h}{t^2}$

The recommended technique is to vary $h$ over a range (say $0.2$ m to $1.5$ m), measure $t$ at each $h$ , and plot $h$ against $t^2$ . The gradient of the best-fit straight line is $g/2$ , so $g = 2 \times \text{gradient}$ .

A typical school-lab result gives $g = 9.8 \pm 0.1$ m s $^{-2}$ .

Principal sources of error:

Human reaction time if using a stopwatch — eliminate with light gates or trapdoor electronic timing.
Air resistance — small for a dense steel ball over a short drop, but not exactly zero; results trend slightly low.
Release delay from the electromagnet — the residual magnetic field takes a few milliseconds to fully decay after the switch opens, contributing a small systematic offset.
Light-gate beam thickness — the ball's diameter is finite, so timing depends on which edge of the ball triggers the beam.

The plot- $h$ -vs- $t^2$ -and-find-gradient method is preferred over computing $g$ at each $h$ and averaging, because it:

(i) automatically handles the systematic offset of release-delay as an intercept on the time axis (which can then be subtracted), and (ii) gives a least-squares optimal estimate, with the random scatter visible as deviations from the line.

Projectile Motion — The Big Idea

A projectile is an object launched with some initial velocity and thereafter moving under gravity alone (air resistance ignored). Examples: a cricket ball, an artillery shell, a long-jumper, a stone thrown from a cliff.

The central physical insight is:

Horizontal and vertical motion are completely independent. They share the same clock, but the horizontal velocity does not influence vertical motion, and vice versa.

Horizontal: no force acts (ignoring air resistance), so $a_x = 0$ and $v_x$ is constant. Use $s_x = v_x t$ .
Vertical: constant acceleration $-g$ downward (with upward positive). Use SUVAT in the form $s_y = u_y t - \tfrac{1}{2} g t^2$ etc.

Galileo arrived at this independence by a beautiful thought experiment: imagine a ball dropped from a moving ship's mast. From the ship's deck the ball appears to fall straight down; from the shore it appears to follow a parabolic arc. Both observations are correct — the horizontal velocity is the ship's velocity (constant), and the vertical motion is independent free fall. The parabolic trajectory you see from shore is the sum of the two.

Horizontal Launch — Worked Example

A ball rolls off a horizontal table of height $1.2$ m with horizontal velocity $3.0$ m s $^{-1}$ . Find (a) the time to hit the floor and (b) the horizontal distance covered.

Vertical (downward positive): $u_y = 0$ , $a = g$ , $s_y = 1.2$ m.

$1.2 = 0 + \tfrac{1}{2} \times 9.81 \times t^2 \Rightarrow t = \sqrt{\frac{1.2}{4.905}} = 0.495 \text{ s}$

Horizontal (constant velocity 3.0 m s $^{-1}$ ):

$s_x = v_x \times t = 3.0 \times 0.495 = 1.48 \text{ m}$

So the ball hits the floor $1.48$ m from the edge of the table, $0.495$ s after rolling off.

Notice: the horizontal velocity ( $3.0$ m s $^{-1}$ ) plays no role in the vertical equations and the gravitational acceleration ( $g$ ) plays no role in the horizontal equation. They are coupled only by the shared time $t$ .

Projectile at an Angle — Worked Example

A football is kicked from ground level at $u = 25$ m s $^{-1}$ at $\theta = 40^{\circ}$ above horizontal. Find (a) maximum height, (b) total time of flight, (c) horizontal range.

Decompose the initial velocity:

$u_x = u \cos \theta = 25 \cos 40^{\circ} = 19.15$ m s $^{-1}$ .
$u_y = u \sin \theta = 25 \sin 40^{\circ} = 16.07$ m s $^{-1}$ .

(a) Maximum height — vertical motion only, $v_y = 0$ at apex:

$0 = u_y^2 - 2gH \Rightarrow H = \frac{u_y^2}{2g} = \frac{258.2}{19.62} = 13.2 \text{ m}$

(b) Time of flight — return to launch height $s_y = 0$ :

$0 = u_y t - \tfrac{1}{2} g t^2 = t(u_y - \tfrac{1}{2} g t) \Rightarrow t = \frac{2 u_y}{g} = \frac{32.14}{9.81} = 3.28 \text{ s}$

(c) Horizontal range:

$R = u_x \times t = 19.15 \times 3.28 = 62.8 \text{ m}$

The Range Equation

For a projectile launched from ground level at angle $\theta$ with speed $u$ , landing at the same height:

$R = u_x \times t_{\text{flight}} = (u \cos \theta) \times \frac{2 u \sin \theta}{g} = \frac{u^2 \sin(2\theta)}{g}$

using the double-angle identity $2 \sin\theta \cos\theta = \sin(2\theta)$ .

$\boxed{R = \frac{u^2 \sin(2\theta)}{g}}$

Maximum range occurs when $\sin(2\theta) = 1$ , i.e. $2\theta = 90^{\circ}$ , so $\theta = 45^{\circ}$ . Any deviation from $45^{\circ}$ reduces the range (for fixed launch speed on level ground). This is why long-jumpers, shot-putters and ski-jumpers all aim for launch angles close to (but, due to air resistance, slightly below) $45^{\circ}$ .

A useful symmetry: $\sin(2\theta) = \sin(2(90^{\circ} - \theta))$ , so $\theta$ and $(90^{\circ} - \theta)$ give the same range for the same launch speed. For example, $\theta = 30^{\circ}$ and $\theta = 60^{\circ}$ yield identical $R$ but very different apex heights and times of flight.

Exam Tip: The range equation only holds when launch and landing heights are equal. If the ground falls away (golf tee on a hillside) or rises (boundary shot to a higher fielder), go back to decomposing velocity and applying SUVAT in $x$ and $y$ separately. Never substitute into $R = u^2 \sin(2\theta) / g$ when launch and landing heights differ.

Worked Example 4 — Launch above Ground

A stone is thrown at $20$ m s $^{-1}$ at $35^{\circ}$ above horizontal from a 12-m-high cliff. Find (a) time of flight to the ground and (b) horizontal range.

Free Fall and Projectile Motion

Free Fall and Projectile Motion

Free Fall — Definitions and Universality

Why mass cancels — the universality of free fall

Vertical Free Fall — Worked Examples

Worked Example 1 — Dropped Stone

Worked Example 2 — Upward Throw and Symmetry

Worked Example 3 — Asymmetric drop from a cliff

PAG 6 — Measuring ggg Experimentally

Projectile Motion — The Big Idea

Horizontal Launch — Worked Example

Projectile at an Angle — Worked Example

The Range Equation

Worked Example 4 — Launch above Ground

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PAG 6 — Measuring $g$ Experimentally