An object launched horizontal from height h: compute fall time\n t=√(2h/g), horizontal range R=v₀·t, impact velocity & angle, and kinetic energy on impact.\n With inverse solver, Earth/Moon/Mars gravity presets, and parabola trajectory diagram.\n Everything computed in your browser; nothing uploaded.
Gravity (applies to all modes)
Nominal surface values as used in physics textbooks.
Mode 1 — From velocity & height
How long airborne, how far it lands, and how hard the impact?
Fall time
0s
Horizontal range R
0m
Impact velocity
0m/s
Impact angle
0°
Vertical impact component
0m/s
Impact kinetic energy
0J
g used: 9.80665 m/s²
Mode 2 — Velocity for target range (inverse)
What v₀ lands exactly R away from height h?
Required initial velocity
0m/s
Mode 3 — Height for target range (inverse)
From what height must you launch with v₀ to land R away?
Ketinggian dibutuhkan
0m
Worked example
Horizontal launch v₀=10 m/s from h=20 m (Earth)
t = √(2·20/9.80665) = 2.0196 s; R = 10·2.0196 = 20.196 m;
vy impact = 19.806 m/s; vimpact = √(10² + 19.806²) = 22.187 m/s;\n impact angle θ = atan(19.806/10) = 63.21° below horizontal;\n KE (m=1 kg) = ½·1·22.187² = 246.13 J.
Other gravities: v₀=10 m/s, h=20 m
On Moon (g=1.62): t = 4.9690 s, R = 49.690 m — floats more than 2× farther.\n On Mars (g=3.71): t = 3.2835 s, R = 32.835 m.
Inverse: target R=30 m with v₀=15 m/s
h = g·R²/(2·v₀²) = 9.80665·900/450 = 19.61 m.\n Reverse check: t = √(2·19.6133/9.80665) = 2.0000 s → R = 15·2 = 30.00 m. ✓
Horizontal and vertical motion are independent: komponen horizontal vx = v₀\n constant (no horizontal force), vertical component accelerated by gravity from zero.\n Hence fall time depends only on h: t = √(2h/g), and v₀ does not affect t at all.\n Inverse uses analytic solutions v₀ = R·√(g/(2h)) and h = g·R²/(2·v₀²), both round-trip validated.\n Honest assumptions: hambatan udara diabaikan (vacuum model, like textbook problems);\n Moon 1.62 and Mars 3.71 are nominal surface values, not the latest scientific precision constants.
Scope: what is (and isn’t) computed
This tool = pure HORIZONTAL launches (0° angle). School-physics context:\n fall time, horizontal range, impact velocity vector, kinetic energy, and planet gravity comparison.
NOT an angled-launch calculator. For projectiles from a ramp at 0–60°\n (car-stunt context), use\n
car-jump-distance (#1555).
NOT a free-fall calculator. For objects dropped vertically with no initial velocity,\n use free-fall-velocity.
NOT a drag-ballistics tool. For projectiles with significant air resistance\n (bullets, model rockets), use\n
ballistic-coefficient.
What is the formula for horizontal projectile motion?
For an object launched horizontally with initial velocity v₀ from height h (no air resistance): fall time t = √(2h/g); horizontal range R = v₀·t; vertical impact velocity v_y = g·t; impact speed v = √(v₀² + v_y²); impact angle θ = atan(v_y/v₀) below horizontal; impact kinetic energy KE = ½·m·v². Verified example: v₀ = 10 m/s, h = 20 m, g = 9.80665 m/s² → t = 2.0196 s; R = 20.196 m; v_y = 19.806 m/s; v_impact = 22.187 m/s; θ = 63.21°; KE (m = 1 kg) = 246.13 J.
Why doesn’t horizontal motion affect fall time?
Because horizontal and vertical motions are independent (Galileo’s principle of motion independence): gravity only pulls downward, so vertical acceleration is always g and horizontal velocity v₀ stays constant (no air resistance). Fall time is set purely by height: t = √(2h/g) — v₀ does not appear in that formula. That is why two objects, one dropped and one fired horizontally from the same height, land simultaneously. This calculator proves it: change v₀ however you like, the fall-time column never changes.
How do you calculate speed and angle when the object hits the ground?
Break velocity into components. The horizontal component stays v_x = v₀ (no horizontal force). The vertical component grows under gravity: v_y = g·t = √(2gh). The resultant v_impact = √(v₀² + v_y²), and the angle below horizontal θ = atan(v_y/v₀). Example: v₀ = 10 m/s from h = 20 m → v_y = 19.806 m/s, v_impact = 22.187 m/s, θ = 63.21° — the object arrives steeper than 45° because its vertical fall outruns its horizontal motion.
How do you find v₀ when range and height are known (inverse problem)?
Use inverse mode 1: v₀ = R/t = R·√(g/(2h)). Example: target landing 20.196 m from the base of a 20 m launch point → v₀ = 10 m/s (round-trip verified: entering v₀ = 10 m/s, h = 20 m gives back R = 20.196 m). This mode suits ‘what throw speed hits a target R away’ type problems.
How do you find height when range and initial velocity are known?
Use inverse mode 2: h = g·R²/(2·v₀²). Example: R = 30 m, v₀ = 15 m/s, g = 9.80665 m/s² → h = 19.61 m (exact 19.6133 m). Reverse check: v₀ = 15 m/s from h = 19.6133 m gives t = 2.0000 s and R = 30.00 m — consistent.
Which gravity values are available? Are they accurate for the Moon and Mars?
Presets: Earth 9.80665 m/s² (international standard value, exact by definition), Moon 1.62 m/s², and Mars 3.71 m/s² — all nominal surface values as used in textbooks, not the latest scientific precision constants, plus a custom option for any value. Example: v₀ = 10 m/s from h = 20 m on the Moon → t = 4.9690 s, R = 49.690 m; on Mars → t = 3.2835 s, R = 32.835 m. Weaker gravity = longer airtime = farther range.
Does this calculator account for air resistance?
No — the model is vacuum, like school physics problems. For low speeds and dense objects the results are close to reality; for light objects/high speeds, air drag shortens range and bends the trajectory earlier, so figures here are a theoretical upper bound. For serious air-drag modeling, see our ballistic-coefficient tool.
What happens in edge cases: h = 0 or v₀ = 0?
Both are handled honestly by the formulas. If h = 0 (launched from ground level), t = 0 and R = 0 — the object ‘lands’ immediately. If v₀ = 0, the object falls purely vertically: R = 0, and its impact speed v = √(2gh) — exactly the free-fall formula. Try inverse mode with h = 0: the calculator refuses because there is no solution (division by zero).
What units are used? Can I use km/h?
This calculator uses pure SI units (m/s, meters, kg, seconds) so the formulas apply directly with no hidden conversion factors: t = √(2h/g) is only valid with h in meters and g in m/s². For km/h, first divide by 3.6 (e.g. 36 km/h = 10 m/s), then enter the result.
How is this tool different from the Car Jump Distance Calculator?
The Car Jump Distance Calculator (#1555) computes projectiles launched from a ramp at an ANGLE (0–60°) — car-stunt context, fixed Earth gravity, no impact speed, impact angle, kinetic energy, planet gravity presets, or trajectory diagram. This tool is for pure HORIZONTAL launches (0° angle): school-physics context focused on horizontal–vertical motion independence, impact velocity vectors, kinetic energy, and Earth–Moon–Mars gravity comparison. Both use the same vacuum projectile physics, but answer different questions.
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