A Rock Is Thrown Horizontally with Speed v: Which Formula, and Why Not the Other One

A rock thrown horizontally with speed v from a height h stays in the air for t = √(2h/g), lands a horizontal distance R = v√(2h/g) from the base of the launch point, and arrives at a speed of √(v² + 2gh). The time of flight contains no v, so the thrown rock hits the ground at the same instant as one simply dropped beside it. Using the standard acceleration of free fall adopted by the 3rd General Conference on Weights and Measures in 1901 (980.665 cm/s², or 9.80665 m/s²), a rock thrown horizontally at 12.0 m/s from a height of 10.00 m falls for 1.4281 s, travels 17.14 m horizontally, and strikes at 18.44 m/s along a path 49.4° below the horizontal. The angled-launch range formula R = v² sin 2θ / g returns zero for this problem and cannot be used, because it assumes the projectile lands at the height it was launched from.

At the rental desk where I learned this job, a rock meant one thing. A curling stone: no heavier than 19.96 kg and no more than 914 mm around, under rule R2(a) of World Curling's Rules of Curling, and it never left the ice. So a physics question with a rock in it triggers the desk reflex before it triggers any algebra. Which rock, thrown from where, and which rule covers it.

That reflex is the whole answer here. Most wrong solutions I see to this problem arrive with clean arithmetic. The student reached for the range equation memorised last week, which was written for a different situation and quietly hands back nonsense in this one.

Why the rock's time in the air ignores speed v entirely

Throwing horizontally means the initial vertical velocity is zero. From the moment of release the rock runs two motions that never speak to each other. It travels sideways at a constant v, and it falls exactly the way a dropped rock falls.

The falling half gives h = ½gt², which rearranges to t = √(2h/g).

Take a height with paperwork behind it. World Aquatics puts its highest diving platform at 10 metres, and its Diving Pool Certification, in force from 1 April 2026, permits the as-built height to vary by plus 0.05 m and minus 0.00 m from that figure. With h = 10.00 m:

t = √(2 × 10.00 / 9.80665) = 1.4281 s

Every rock thrown horizontally off that platform is airborne for 1.4281 s. A JUGS BP1 pitching machine, whose spec sheet gives a dial range of 15 to 70 mph, would launch one at 6.71 m/s at the bottom of its range and 31.29 m/s at the top. Both land 1.4281 s after release. Only the landing point moves.

Students distrust that result more than anything else in the chapter, and the reason is worth naming. Gravity has no way to know about the sideways motion. It pulls down on a fast rock exactly as hard as on a slow one, and it has the same 10.00 m to work through either way.

How far from the base the rock lands

Horizontal speed never changes in the idealised problem, so the range is just speed multiplied by the time you already found:

R = v t = v√(2h/g)

Range is linear in v. Double the throw and you double the distance, with no square anywhere.

At 12.0 m/s from 10.00 m, R = 12.0 × 1.4281 = 17.14 m.

I picked 12.0 m/s because it converts to 26.8 mph, which sits comfortably inside the BP1's dial. It is a speed you can set, read off the machine's digital display, and reproduce, rather than a number chosen to make the answer come out round.

| Speed v | Range R from 10.00 m | Impact speed | Angle below horizontal | |---|---|---|---| | 6.71 m/s (15 mph, BP1 minimum) | 9.58 m | 15.53 m/s | 64.4° | | 8.94 m/s (20 mph) | 12.77 m | 16.62 m/s | 57.4° | | 9.45 m/s (21.1 mph) | 13.50 m | 16.90 m/s | 56.0° | | 12.0 m/s (26.8 mph) | 17.14 m | 18.44 m/s | 49.4° | | 13.41 m/s (30 mph) | 19.15 m | 19.39 m/s | 46.2° | | 31.29 m/s (70 mph, BP1 maximum) | 44.69 m | 34.28 m/s | 24.1° |

The 13.50 m row is in there deliberately. World Aquatics requires 13.50 m of clear water ahead of the plummet for a 10 metre platform, listed as dimension D-10 in the appendix tables to Part 4 of its competition regulations. A rock leaving that platform at 9.45 m/s, about 21 mph, lands exactly at the far wall of a minimum-compliant pool. Anything faster clears the water entirely. The rulebook, without meaning to, has published the answer to a projectile problem.

How to solve any horizontal-launch problem in four steps

1. Write down h, v and g, and set the initial vertical velocity to zero. This is what "thrown horizontally" buys you.
2. Solve the vertical motion on its own for time: t = √(2h/g). Ignore v completely at this stage.
3. Multiply that time by the horizontal speed for the range: R = v t.
4. Build the impact velocity from its two components: the horizontal one is still v, the vertical one is v_y = gt, and the speed is √(v² + v_y²).

How fast the rock is moving when it hits

The horizontal component stays at v for the entire flight. The vertical component grows from zero as v_y = gt, which at the moment of landing equals √(2gh). From 10.00 m that vertical component is 14.00 m/s, no matter what v was.

The impact speed is the hypotenuse of those two:

v_impact = √(v² + 2gh)

For v = 12.0 m/s and h = 10.00 m, that is √(144 + 196.13) = 18.44 m/s. The arrival angle comes from the same triangle, arctan(14.00 / 12.0) = 49.4° below horizontal.

The error I see most often is adding the two components straight, which would give 26.00 m/s here against a true 18.44. Velocities at right angles combine through Pythagoras, and the gap between 26.00 and 18.44 is large enough that it should set off an alarm on its own.

Why the R = v² sin 2θ / g formula returns zero here

OpenStax's University Physics gives the range equation as R = v₀² sin 2θ₀ / g, numbered 4.26 in section 4.3, and attaches an explicit condition to it: the equation holds only for launch and impact on a horizontal surface.

Feed a horizontal throw into it. θ = 0, sin 0 = 0, so R = 0.

The formula is being honest about its own assumption. A projectile launched flat from ground level, landing at ground level, has travelled nowhere, because it was already on the ground when it started. Your rock is not on the ground. It begins 10 metres above where it finishes, and there is no h anywhere in that equation to describe the difference.

This is why the horizontal launch gets its own pair of equations instead of being treated as a special case of the angled one. The launch height enters through the time of flight, and time of flight is precisely the quantity the angled range formula has already solved away and discarded.

Horizontal launch and angled launch, run on identical numbers

Same rock, same 12.0 m/s, same g = 9.80665 m/s². Only the angle and the starting height change.

| Launch | Time of flight | Range | Apex above launch | Impact speed | Impact angle | |---|---|---|---|---|---| | Horizontal, from 10.00 m | 1.43 s | 17.14 m | 0 m | 18.44 m/s | 49.4° | | 33.05° up, from 10.00 m | 2.24 s | 22.57 m | 2.18 m | 18.44 m/s | 57.0° | | 45° up, from 10.00 m | 2.54 s | 21.51 m | 3.67 m | 18.44 m/s | 62.6° | | 60° up, from 10.00 m | 2.84 s | 17.03 m | 5.51 m | 18.44 m/s | 71.0° | | 45° up, from ground level | 1.73 s | 14.68 m | 3.67 m | 12.00 m/s | 45.0° |

Three results fall out of that table, and each one is quotable on its own.

Impact speed is 18.44 m/s in every row launched from 10.00 m, whatever the angle. Energy does not track direction, only the height you fell through and the speed you started with. What the angle changes is where the rock lands and how steeply it arrives.

Forty-five degrees stops being optimal the moment you launch from a height. The best angle from 10.00 m at this speed is 33.05°, given by arctan(v / √(v² + 2gh)), and it buys 22.57 m. That optimum slides toward zero as the launch height grows, which is why artillery and cliff-edge problems never sit at 45°. Push past the optimum and you lose ground fast: 60° from the same platform returns 17.03 m, fractionally short of simply letting the rock go flat.

The flat throw from 10 metres beats the textbook-perfect 45° throw from flat ground by 2.45 m at the same 12.0 m/s. Height outbids angle here. The crossover sits at h = v²/(2g), which for 12.0 m/s works out to 7.34 m. Launch from higher than that and the horizontal throw wins, every time, against the best possible ground-level angle.

Which value of g to use, and whether your location changes the answer

This is my desk question, transplanted into a physics chapter. Readers ask me more often than anything else which local rule changes the answer. So I went and got the local rule.

The 9.80665 m/s² printed in every textbook is a convention rather than a measurement of your town. The 3rd General Conference on Weights and Measures fixed it in 1901, in a one-sentence resolution: the value adopted in the International Service of Weights and Measures for the standard acceleration due to gravity is 980.665 cm/s². It is a number by agreement, chosen so that everyone's calculations agree with everyone else's.

Real surface gravity gets measured. NOAA's National Geodetic Survey will predict it for any coordinates in its database, interpolated from observed values referenced to the International Gravity Standardization Net 1971. I ran three US locations through it.

| Location | Coordinates and MSL height | NGS predicted gravity | g in m/s² | |---|---|---|---| | Key West, Florida | 24.5551° N, 81.7800° W, 2 m | 978,955 ± 2 mGal | 9.78955 | | Denver, Colorado | 39.7392° N, 104.9903° W, 1609 m | 979,612 ± 2 mGal | 9.79612 | | Anchorage, Alaska | 61.2181° N, 149.9003° W, 31 m | 981,918 ± 4 mGal | 9.81918 |

The spread from the Florida Keys to south-central Alaska is 0.0296 m/s², about 0.30%. Run the same 12.0 m/s throw off the same 10.00 m platform in each place and the rock lands 17.152 m out in Key West against 17.126 m in Anchorage. The entire continental variation in gravity moves the landing point 26 millimetres.

Set that beside the platform itself. World Aquatics allows a certified 10 metre platform to be built up to 0.05 m high. That tolerance alone shifts the landing point by 43 mm, which is more than the gravity difference between Florida and Alaska.

Use 9.81 m/s², or 9.8 if that is what your instructor writes on the board, and stop worrying about it. The local rule is real and it is measurable, and it is worth less than the tolerance on the concrete you are standing on. The formula choice is what moves your answer by metres.

What the clean answer leaves out

Air resistance, first. Every figure above assumes a vacuum, which a diving pool deck is not. For a dense rock over a 10 metre drop the neglected drag is small, and for a light one it is not small at all. I have not measured it and I will not invent a correction factor. What I can point at is where the sensitivity lives: drag grows with the square of speed and with frontal area divided by mass, so the 70 mph row in the first table is far more wrong than the 15 mph row.

Mass never appears in any equation on this page, which reliably surprises people. A 19.96 kg curling stone at the World Curling maximum and a pebble, released at the same 12.0 m/s from the same platform, land in the same place at the same instant in the idealised problem. Mass drops out of h = ½gt² because the mass resisting acceleration is the same mass being pulled. Restore the air and the stone wins, since it carries far more inertia behind each square centimetre of frontal area.

One more thing the problem statement usually hides. The h you want is measured from the release point to the landing surface, not from the thrower's feet and not from the deck. Someone standing at the front edge of a certified 10 metre platform releases a rock at roughly shoulder height, call it 1.40 m above the deck, which makes h = 11.40 m. Time rises to 1.5248 s and the range to 18.30 m. That is 1.16 m further out than the deck-height answer, forty-five times the entire Key West to Anchorage gravity effect, from a detail most solutions never mention.

Frequently asked questions

What is the formula when a rock is thrown horizontally with speed v?

Three equations cover it. Time of flight is t = √(2h/g), horizontal range is R = v√(2h/g), and impact speed is √(v² + 2gh), where h is the launch height above the landing surface and g is 9.80665 m/s². The rock's mass is not needed.

Does throwing the rock faster make it hit the ground sooner?

No. Time of flight depends only on the launch height and gravity, through t = √(2h/g). A rock thrown at 6.71 m/s and one thrown at 31.29 m/s from the same 10.00 m height both land 1.4281 s after release. The faster rock simply lands further away.

How do I find the time for a stone thrown horizontally from height h?

Use the vertical motion alone. Throwing horizontally sets the initial vertical velocity to zero, so h = ½gt², which rearranges to t = √(2h/g). Do this before touching the horizontal speed. From 10.00 m with g = 9.80665 m/s², the time is 1.4281 seconds.

Why does R = v² sin 2θ / g give zero for a horizontal throw?

Because sin 0 = 0. That formula assumes the projectile lands at the same height it was launched from, a condition OpenStax states explicitly alongside the equation. A flat throw from ground level lands immediately, so zero is correct for its assumptions and useless for a cliff or a platform.

How fast is the rock going when it hits the ground?

Combine the two velocity components with Pythagoras: √(v² + 2gh). At 12.0 m/s from 10.00 m that gives 18.44 m/s, arriving 49.4° below horizontal. Adding the components directly would give 26.00 m/s, which is wrong. The impact speed is identical for any launch angle at that speed and height.

Should I use 9.8 or 9.81 for g in projectile problems?

Either. The difference is negligible for homework. Standard gravity is 9.80665 m/s², fixed by international agreement in 1901, while measured local values across the United States run from about 9.7896 to 9.8192. That whole spread changes a 17-metre range by 26 millimetres.

Does the mass of the rock matter in a projectile problem?

Not in the idealised version. Mass cancels out of every equation, so a 19.96 kg curling stone and a pebble thrown at the same speed from the same height land together. Mass starts to matter once air resistance is included, because heavier objects of the same size decelerate less.