{"id":226,"date":"2026-06-13T16:53:14","date_gmt":"2026-06-13T16:53:14","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=226"},"modified":"2026-06-13T16:53:16","modified_gmt":"2026-06-13T16:53:16","slug":"terminal-velocity","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/terminal-velocity\/","title":{"rendered":"What Is Terminal Velocity?"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\n\nTerminal velocity is the constant maximum speed a falling object reaches when the upward drag force from the air exactly balances its downward weight, leaving zero net force and zero acceleration. It depends on the object&#8217;s mass, cross-sectional area, shape and the air&#8217;s density \u2014 not on how far the object has already fallen.\n\n<\/p><\/div>\n<p>Step out of a plane and, for the first few seconds, you really do accelerate \u2014 the ground rushes up faster and faster. Then something odd happens. The rushing stops getting worse: you are still plummeting, yet your speed locks in and refuses to climb any higher.<\/p>\n<p>That ceiling is terminal velocity. It is why a skydiver, a hailstone and a falling leaf each settle into a steady descent instead of speeding up without limit, and it is the reason a parachute can save your life. The whole idea comes down to one tug-of-war: gravity pulling down, air pushing back.<\/p>\n<h2>What Is Terminal Velocity?<\/h2>\n<p>Picture two forces fighting over a falling object. Gravity pulls it down with a steady force \u2014 its weight. The air shoves back with a drag force that points up, against the motion. The catch is that drag is not constant: the faster you move, the harder the air resists.<\/p>\n<p>At the instant of release the object is barely moving, so drag is almost nothing and gravity wins easily. As speed builds, drag grows quickly. Eventually the upward drag matches the downward weight exactly.<\/p>\n<p>When those two forces cancel, the <strong>net force is zero<\/strong>. By Newton&#8217;s first law, an object with no net force keeps moving at a constant velocity \u2014 it stops accelerating. That steady, top speed is the terminal velocity. The object is still falling fast; it simply cannot go any faster in those conditions.<\/p>\n<svg viewBox=\"0 0 720 400\" role=\"img\" aria-label=\"Three stages of a falling object showing the weight arrow staying constant while the drag arrow grows until it equals weight at terminal velocity\" style=\"width:100%;height:auto;max-width:720px;display:block;margin:24px auto;\">\n  <rect x=\"2\" y=\"2\" width=\"716\" height=\"396\" rx=\"12\" fill=\"#F5F2EA\" stroke=\"#D9CFB8\" stroke-width=\"1.5\"><\/rect>\n  <text x=\"360\" y=\"38\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"18\" font-weight=\"700\" fill=\"#0A1628\">How the forces change as an object falls<\/text>\n  <!-- Stage 1: x=160 -->\n  <circle cx=\"160\" cy=\"160\" r=\"15\" fill=\"#142139\" stroke=\"#C8932A\" stroke-width=\"2.5\"><\/circle>\n  <line x1=\"194\" y1=\"158\" x2=\"194\" y2=\"256\" stroke=\"#1F2E47\" stroke-width=\"3\"><\/line>\n  <polygon points=\"188,244 200,244 194,260\" fill=\"#1F2E47\"><\/polygon>\n  <text x=\"194\" y=\"280\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#1F2E47\">mg<\/text>\n  <text x=\"118\" y=\"164\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12\" font-weight=\"600\" fill=\"#7A1F2B\">Fd \u2248 0<\/text>\n  <text x=\"160\" y=\"312\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" font-weight=\"700\" fill=\"#0A1628\">1 \u00b7 Just released<\/text>\n  <text x=\"160\" y=\"334\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12.5\" fill=\"#4A5A72\">drag = 0, a = g<\/text>\n  <!-- Stage 2: x=360 -->\n  <circle cx=\"360\" cy=\"160\" r=\"15\" fill=\"#142139\" stroke=\"#C8932A\" stroke-width=\"2.5\"><\/circle>\n  <line x1=\"394\" y1=\"158\" x2=\"394\" y2=\"256\" stroke=\"#1F2E47\" stroke-width=\"3\"><\/line>\n  <polygon points=\"388,244 400,244 394,260\" fill=\"#1F2E47\"><\/polygon>\n  <text x=\"394\" y=\"280\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#1F2E47\">mg<\/text>\n  <line x1=\"326\" y1=\"158\" x2=\"326\" y2=\"113\" stroke=\"#C8932A\" stroke-width=\"3\"><\/line>\n  <polygon points=\"320,121 332,121 326,107\" fill=\"#C8932A\"><\/polygon>\n  <text x=\"326\" y=\"100\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12\" font-weight=\"700\" fill=\"#C8932A\">Fd<\/text>\n  <text x=\"360\" y=\"312\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" font-weight=\"700\" fill=\"#0A1628\">2 \u00b7 Speeding up<\/text>\n  <text x=\"360\" y=\"334\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12.5\" fill=\"#4A5A72\">drag &lt; weight, a shrinking<\/text>\n  <!-- Stage 3: x=560 -->\n  <circle cx=\"560\" cy=\"160\" r=\"15\" fill=\"#142139\" stroke=\"#C8932A\" stroke-width=\"2.5\"><\/circle>\n  <line x1=\"594\" y1=\"158\" x2=\"594\" y2=\"256\" stroke=\"#1F2E47\" stroke-width=\"3\"><\/line>\n  <polygon points=\"588,244 600,244 594,260\" fill=\"#1F2E47\"><\/polygon>\n  <text x=\"594\" y=\"280\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#1F2E47\">mg<\/text>\n  <line x1=\"526\" y1=\"158\" x2=\"526\" y2=\"60\" stroke=\"#C8932A\" stroke-width=\"3\"><\/line>\n  <polygon points=\"520,68 532,68 526,54\" fill=\"#C8932A\"><\/polygon>\n  <text x=\"526\" y=\"47\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12\" font-weight=\"700\" fill=\"#C8932A\">Fd<\/text>\n  <text x=\"560\" y=\"312\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" font-weight=\"700\" fill=\"#0A1628\">3 \u00b7 Terminal velocity<\/text>\n  <text x=\"560\" y=\"334\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12.5\" fill=\"#4A5A72\">drag = weight, a = 0<\/text>\n<text x=\"360\" y=\"372\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12\" font-style=\"italic\" fill=\"#7A1F2B\">Weight (mg) stays fixed; drag (Fd) grows with speed until the two arrows match.<\/text>\n\n<\/svg>\n<p style=\"text-align:center;font-size:13px;color:#4A5A72;font-style:italic;margin-top:-6px;\">Figure 1: A falling object reaches terminal velocity the moment the growing upward drag equals its constant downward weight.<\/p>\n<p>Notice what terminal velocity is <em>not<\/em>. It is not the speed of a normal free fall in a vacuum, where nothing pushes back. It needs a fluid \u2014 air, water, oil \u2014 to exist at all.<\/p>\n<h2>The Terminal Velocity Formula<\/h2>\n<p>Two things set the answer: how hard gravity pulls (the weight) and how hard the air pushes (the drag). Write down the drag force, set it equal to the weight, and the speed that makes them balance drops straight out.<\/p>\n<div class=\"pf-formula\">v\u209c = \u221a( 2mg \/ (\u03c1 \u00b7 C_d \u00b7 A) )<\/div>\n<p>Here each symbol means:<\/p>\n<ul>\n<li><strong>v\u209c<\/strong> \u2014 terminal velocity, in metres per second (m\/s).<\/li>\n<li><strong>m<\/strong> \u2014 mass of the object, in kilograms (kg).<\/li>\n<li><strong>g<\/strong> \u2014 gravitational field strength, \u2248 9.81 m\/s\u00b2 near Earth&#8217;s surface.<\/li>\n<li><strong>\u03c1<\/strong> (rho) \u2014 density of the fluid; for air at sea level and 15 \u00b0C, \u03c1 \u2248 1.225 kg\/m\u00b3.<\/li>\n<li><strong>A<\/strong> \u2014 cross-sectional (frontal) area facing the flow, in square metres (m\u00b2).<\/li>\n<li><strong>C_d<\/strong> \u2014 drag coefficient, a dimensionless number fixed by the object&#8217;s shape and surface (\u2248 0.47 for a smooth sphere, \u2248 1.0\u20131.3 for a flat plate).<\/li>\n<\/ul>\n<p>The formula comes from the standard drag equation, which describes the resisting force of the air at everyday falling speeds:<\/p>\n<div class=\"pf-formula\">F_d = \u00bd \u00b7 \u03c1 \u00b7 v\u00b2 \u00b7 C_d \u00b7 A[\/pf_formula&gt;\n<p>At terminal velocity the drag force equals the weight, so F_d = mg. Substitute mg for F_d, then rearrange for v, and you arrive at the terminal-velocity formula above. NASA derives exactly the same result on its <a href=\"https:\/\/www.grc.nasa.gov\/www\/k-12\/VirtualAero\/BottleRocket\/airplane\/termv.html\" target=\"_blank\" rel=\"noopener\">Terminal Velocity page<\/a>.<\/p>\n<p>Read the formula and the behaviour makes sense. More mass on top pushes v\u209c up. More area or a higher drag coefficient on the bottom pulls v\u209c down. That is why a feather, all area and almost no mass, drifts, while a steel ball bearing of the same size plummets.<\/p>\n<h3>The low-speed case: Stokes&#8217; law<\/h3>\n<p>The square-of-speed drag above applies to skydivers, raindrops and most falling objects you can see. For very small, very slow objects \u2014 mist, dust, microscopic droplets \u2014 the air behaves differently and drag becomes proportional to speed itself. Terminal velocity then follows Stokes&#8217; law:<\/p>\n[pf_formula]v\u209c = 2r\u00b2(\u03c1_p \u2212 \u03c1_f) g \/ (9\u03b7)<\/div>\n<ul>\n<li><strong>r<\/strong> \u2014 radius of the (spherical) particle, in metres (m).<\/li>\n<li><strong>\u03c1_p<\/strong> \u2014 density of the particle, in kg\/m\u00b3.<\/li>\n<li><strong>\u03c1_f<\/strong> \u2014 density of the fluid, in kg\/m\u00b3.<\/li>\n<li><strong>\u03b7<\/strong> (eta) \u2014 dynamic viscosity of the fluid; for air, \u03b7 \u2248 1.8 \u00d7 10\u207b\u2075 Pa\u00b7s.<\/li>\n<\/ul>\n<p>This is why fog hangs in the air for hours. A tiny droplet&#8217;s terminal velocity is only a few centimetres per second, so it barely settles at all. Drag and air resistance are close cousins of ordinary <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/what-is-friction\/\">friction<\/a> \u2014 a force that opposes motion and depends on how fast you move through the fluid.<\/p>\n<h2>How Terminal Velocity Works<\/h2>\n<p>Think of the fall in three acts.<\/p>\n<p><strong>Act one \u2014 release.<\/strong> Speed is zero, so drag is zero. The only force is weight, and the object accelerates at the full value of g (about 9.81 m\/s\u00b2 on Earth). This first moment looks exactly like free fall.<\/p>\n<p><strong>Act two \u2014 speeding up.<\/strong> As the object gains speed, drag climbs steeply, because it grows with the square of speed. The upward drag eats into the downward weight, the net force shrinks, and so does the acceleration. The object is still getting faster, but less and less eagerly.<\/p>\n<p><strong>Act three \u2014 balance.<\/strong> Drag finally equals weight. Net force hits zero, acceleration hits zero, and the speed holds steady. Welcome to terminal velocity.<\/p>\n<p>Strictly, the object approaches this speed without ever quite touching it \u2014 the curve flattens but never becomes perfectly horizontal. In practice that distinction does not matter. A belly-down skydiver reaches about 99% of terminal velocity in roughly 10\u201312 seconds, after falling around 450 m (about 1,500 ft).<\/p>\n<svg viewBox=\"0 0 720 380\" role=\"img\" aria-label=\"Velocity against time graph for a falling object, rising steeply then flattening towards a horizontal dashed line labelled terminal velocity\" style=\"width:100%;height:auto;max-width:720px;display:block;margin:24px auto;\">\n  <rect x=\"2\" y=\"2\" width=\"716\" height=\"376\" rx=\"12\" fill=\"#F5F2EA\" stroke=\"#D9CFB8\" stroke-width=\"1.5\"><\/rect>\n  <text x=\"360\" y=\"36\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"17\" font-weight=\"700\" fill=\"#0A1628\">Speed against time as an object falls<\/text>\n  <!-- axes -->\n  <line x1=\"92\" y1=\"70\" x2=\"92\" y2=\"312\" stroke=\"#1F2E47\" stroke-width=\"2\"><\/line>\n  <line x1=\"92\" y1=\"312\" x2=\"664\" y2=\"312\" stroke=\"#1F2E47\" stroke-width=\"2\"><\/line>\n  <!-- asymptote -->\n  <line x1=\"92\" y1=\"98\" x2=\"664\" y2=\"98\" stroke=\"#C8932A\" stroke-width=\"1.8\" stroke-dasharray=\"6 5\"><\/line>\n  <text x=\"648\" y=\"90\" text-anchor=\"end\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#C8932A\">terminal velocity (v\u209c)<\/text>\n  <!-- curve -->\n  <path d=\"M92,312 C 150,170 230,118 340,104 S 520,99 664,98\" fill=\"none\" stroke=\"#7A1F2B\" stroke-width=\"3.5\"><\/path>\n  <!-- annotations -->\n  <text x=\"168\" y=\"250\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12.5\" font-weight=\"600\" fill=\"#1F2E47\">steep at first<\/text>\n  <text x=\"168\" y=\"268\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12\" fill=\"#4A5A72\">(a = g)<\/text>\n  <text x=\"470\" y=\"128\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12.5\" font-weight=\"600\" fill=\"#1F2E47\">levels off<\/text>\n  <text x=\"470\" y=\"146\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12\" fill=\"#4A5A72\">(a \u2192 0)<\/text>\n  <!-- axis labels -->\n  <text x=\"378\" y=\"344\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"600\" fill=\"#1F2E47\">time (t)<\/text>\n  <text x=\"52\" y=\"195\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"600\" fill=\"#1F2E47\" transform=\"rotate(-90 52 195)\">speed (v)<\/text>\n<\/svg>\n<p style=\"text-align:center;font-size:13px;color:#4A5A72;font-style:italic;margin-top:-6px;\">Figure 2: Speed rises quickly at first, then bends over and approaches a flat ceiling \u2014 the terminal velocity.<\/p>\n<p>Use the interactive lab below to feel this for yourself. Change the mass, frontal area and drag coefficient, drop the object, and watch the drag arrow grow until it cancels the weight and the speed locks in.<\/p>\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Terminal Velocity Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:600px}@media(max-width:760px){.pf-sim-frame{height:1000px}}<\/style><iframe src=\"\/labs\/terminal-velocity.html\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n<h2>Real-World Examples of Terminal Velocity<\/h2>\n<p>You have met terminal velocity more often than you think \u2014 every time it rains.<\/p>\n<p><strong>Raindrops.<\/strong> A tiny drizzle drop about 0.5 mm across falls at only a couple of metres per second, while a large raindrop near 5 mm reaches roughly 9 m\/s (about 20 mph). That is why a downpour stings and a mist does not: bigger drops have a higher terminal velocity. Meteorologists even use the relationship to estimate drop size from how fast rain falls.<\/p>\n<p><strong>Skydivers.<\/strong> A human in the stable belly-to-earth position tops out around 53 m\/s \u2014 roughly 120 mph or 190 km\/h. Tuck into a head-down dive and you cut your frontal area, so drag drops and terminal velocity climbs to 240\u2013290 km\/h (150\u2013180 mph). It is the clearest everyday proof that posture, not just mass, sets your falling speed.<\/p>\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\">\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/06\/CFest18-09750-1200x800-11.jpg\"\n       alt=\"Skydiver at terminal velocity in a belly-to-earth freefall position\"\n       loading=\"lazy\"\n       style=\"width:100%;height:auto;border-radius:4px;\" \/>\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">A belly-to-earth skydiver falls at a terminal velocity of about 120 mph (53 m\/s).<\/figcaption>\n<\/figure>\n<p><strong>The fastest human freefall.<\/strong> In 2012, Felix Baumgartner jumped from about 39 km up as part of the Red Bull Stratos project. He hit a top speed of 1,357.6 km\/h (843.6 mph), Mach 1.25 \u2014 the first person to break the sound barrier in freefall. He could only reach it because the air that high is almost a vacuum, so drag was tiny. Lower down, as the air thickened, his terminal velocity fell back to ordinary skydiving speeds.<\/p>\n<p><strong>Parachutes.<\/strong> A parachute is a terminal-velocity machine. By multiplying the frontal area enormously, it slashes the terminal velocity from a lethal 50-plus m\/s to a survivable few metres per second \u2014 a gentle jog at touchdown rather than a fatal impact.<\/p>\n<p><strong>The Moon.<\/strong> There is no terminal velocity on the Moon. With effectively no atmosphere, there is no drag to balance gravity, so a falling object simply accelerates the whole way down. Apollo 15 astronaut David Scott showed this in 1971 by dropping a hammer and a feather together \u2014 both hit the lunar surface at the same instant, exactly as Galileo predicted for a world without air.<\/p>\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr style=\"background:#142139;color:#FAF6EE;\">\n<th style=\"padding:11px 13px;text-align:left;border:1px solid #D9CFB8;\">Falling object<\/th>\n<th style=\"padding:11px 13px;text-align:left;border:1px solid #D9CFB8;\">Approximate terminal velocity<\/th>\n<th style=\"padding:11px 13px;text-align:left;border:1px solid #D9CFB8;\">Set mainly by<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">Mist \/ fog droplet (~20 \u00b5m)<\/td>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">~0.05 m\/s<\/td>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">Tiny size, viscous (Stokes) drag<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">Drizzle drop (~0.5 mm)<\/td>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">~2 m\/s<\/td>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">Small mass-to-area ratio<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">Large raindrop (~5 mm)<\/td>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">~9 m\/s (\u224820 mph)<\/td>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">Bigger drops fall faster<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">Human skydiver, belly-to-earth<\/td>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">~53 m\/s (\u2248120 mph \/ 190 km\/h)<\/td>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">Large frontal area, high drag<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">Human skydiver, head-down<\/td>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">~67\u201380 m\/s (\u2248150\u2013180 mph)<\/td>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">Streamlined, smaller area<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">Felix Baumgartner (2012, thin stratosphere)<\/td>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">~377 m\/s (1,357.6 km\/h, Mach 1.25)<\/td>\n<td style=\"padding:10px 13px;border:1px solid #D9CFB8;\">Near-vacuum air, almost no drag<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p style=\"font-size:13px;color:#555;\">Values are approximate and shift with posture, clothing, altitude and air density.<\/p>\n<h2>Common Misconceptions About Terminal Velocity<\/h2>\n<p>A few sturdy myths cling to this topic. Here are the four worth clearing up.<\/p>\n<h3>&#8220;In air, all objects fall at the same rate.&#8221;<\/h3>\n<p>Only in a vacuum. Galileo&#8217;s famous rule \u2014 everything falls together \u2014 holds when there is no air resistance. In real air, drag depends on shape and area, so a crumpled ball of paper easily outruns the same sheet held flat. They have identical mass but very different terminal velocities.<\/p>\n<h3>&#8220;Terminal velocity is a fixed number for an object.&#8221;<\/h3>\n<p>It changes with conditions. A skydiver doubles or halves their speed just by changing posture, and terminal velocity also drops as you fall into denser air lower down. Baumgartner&#8217;s record speed existed only in the thin upper atmosphere; it could never happen at sea level.<\/p>\n<h3>&#8220;At terminal velocity there is no force on the object.&#8221;<\/h3>\n<p>Two large forces are still acting \u2014 they simply cancel. Weight pulls down, drag pushes up with equal strength, so the <em>net<\/em> force is zero. Zero net force is not the same as zero force. That is exactly why the speed stays constant rather than dropping.<\/p>\n<h3>&#8220;A heavier object always has a higher terminal velocity.&#8221;<\/h3>\n<p>Only when shape and size are identical. For two matching spheres, the heavier one does fall faster (v\u209c grows with the square root of mass). But across different shapes, area and drag coefficient dominate \u2014 a heavy parachutist falls far slower than a light pebble.<\/p>\n<h2>How Terminal Velocity Relates to Newton&#8217;s Laws, Drag and Free Fall<\/h2>\n<p>Terminal velocity is really just Newton&#8217;s laws in disguise. The balance of forces is pure <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-second-law\/\">Newton&#8217;s second law<\/a>: F = ma. When drag equals weight, the net force F is zero, so the acceleration a must be zero too \u2014 and zero acceleration means a steady velocity, which is <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-laws-of-motion\/\">Newton&#8217;s first law<\/a> in action.<\/p>\n<p>The upward force is drag, a resistive force closely related to ordinary <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/what-is-friction\/\">friction<\/a> but produced by pushing through a fluid. It is the same family of &#8220;opposing&#8221; force, scaled up by speed.<\/p>\n<p>It also sharpens the idea of <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/kinematics\/velocity-vs-speed\/\">velocity versus speed<\/a>. Terminal velocity is a vector \u2014 it has a direction (straight down) as well as a magnitude \u2014 but in a vertical fall its size is simply the steady falling speed.<\/p>\n<p>Finally, it is the missing piece in <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/projectile-motion-guide\/\">projectile motion<\/a>. Most projectile problems assume no air resistance to keep the maths clean. Terminal velocity is what you get when you stop ignoring the air \u2014 the real ceiling that a long fall would actually hit.<\/p>\n<h2>Worked Problems<\/h2>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A 75 kg skydiver falls belly-to-earth with a drag coefficient C_d = 0.70 and a frontal area A = 0.70 m\u00b2. Taking air density \u03c1 = 1.225 kg\/m\u00b3 and g = 9.81 m\/s\u00b2, find their terminal velocity.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong><br>\n\nStep 1: Use v\u209c = \u221a(2mg \/ (\u03c1 \u00b7 C_d \u00b7 A)).<br>\n\nStep 2: Numerator = 2 \u00d7 75 \u00d7 9.81 = 1471.5 ; denominator = 1.225 \u00d7 0.70 \u00d7 0.70 = 0.600.<br>\n\nStep 3: v\u209c = \u221a(1471.5 \/ 0.600) = \u221a2451.5 = 49.5 m\/s.<br>\n\n<strong>Answer: v\u209c \u2248 50 m\/s (about 180 km\/h).<\/strong> Sanity check: real belly-to-earth skydivers reach ~53 m\/s (120 mph); the small difference comes from the exact drag coefficient, area and clothing.\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">For the same 75 kg skydiver, what is the drag force acting on them once they have reached terminal velocity?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong><br>\n\nStep 1: At terminal velocity the net force is zero, so drag exactly equals weight: F_d = mg.<br>\n\nStep 2: F_d = 75 \u00d7 9.81 = 735.75 N.<br>\n\nStep 3: Check with the drag equation: \u00bd \u00d7 1.225 \u00d7 49.5\u00b2 \u00d7 0.70 \u00d7 0.70 \u2248 736 N \u2014 it matches.<br>\n\n<strong>Answer: F_d \u2248 736 N upward, equal and opposite to the weight.<\/strong>\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">The skydiver spreads out and a small drogue increases their effective frontal area by a factor of 4, with everything else unchanged. By what factor does their terminal velocity change?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong><br>\n\nStep 1: From the formula, v\u209c is proportional to 1\/\u221aA (area sits under the square root).<br>\n\nStep 2: Multiplying A by 4 multiplies v\u209c by 1\/\u221a4 = 1\/2.<br>\n\nStep 3: Their terminal velocity halves \u2014 from ~50 m\/s to ~25 m\/s.<br>\n\n<strong>Answer: terminal velocity drops to one half (\u224825 m\/s).<\/strong> This is the principle a parachute uses on a much larger scale.\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">Two spheres have identical size, shape and drag coefficient, but sphere B has 9 times the mass of sphere A. Compare their terminal velocities.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong><br>\n\nStep 1: With shape and area identical, v\u209c is proportional to \u221am.<br>\n\nStep 2: Mass ratio is 9, so the velocity ratio is \u221a9 = 3.<br>\n\nStep 3: Sphere B&#8217;s terminal velocity is 3 times sphere A&#8217;s.<br>\n\n<strong>Answer: B falls 3\u00d7 faster than A.<\/strong> For matching shapes, heavier really does mean faster \u2014 but only because area is held fixed.\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">A table-tennis ball (mass 2.7 g = 0.0027 kg, diameter 40 mm) reaches a terminal velocity of about 9.0 m\/s. Using A = \u03c0r\u00b2, \u03c1 = 1.225 kg\/m\u00b3 and g = 9.81 m\/s\u00b2, find its drag coefficient.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong><br>\n\nStep 1: Rearrange the formula for C_d: C_d = 2mg \/ (\u03c1 \u00b7 A \u00b7 v\u209c\u00b2).<br>\n\nStep 2: A = \u03c0 \u00d7 (0.020)\u00b2 = 1.257 \u00d7 10\u207b\u00b3 m\u00b2 ; numerator = 2 \u00d7 0.0027 \u00d7 9.81 = 0.05297.<br>\n\nStep 3: C_d = 0.05297 \/ (1.225 \u00d7 1.257 \u00d7 10\u207b\u00b3 \u00d7 9.0\u00b2) = 0.05297 \/ 0.1247 = 0.425.<br>\n\n<strong>Answer: C_d \u2248 0.42<\/strong>, close to the textbook value of \u22480.47 for a smooth sphere \u2014 a good confirmation of the drag model.\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A fine water droplet of radius 20 \u00b5m (2.0 \u00d7 10\u207b\u2075 m) falls through still air. Using Stokes&#039; law with water density \u03c1_p = 1000 kg\/m\u00b3, air viscosity \u03b7 = 1.8 \u00d7 10\u207b\u2075 Pa\u00b7s and g = 9.81 m\/s\u00b2 (neglecting air density), find its terminal velocity.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong><br>\n\nStep 1: Use v\u209c = 2r\u00b2\u03c1_p g \/ (9\u03b7) for the low-speed (viscous) regime.<br>\n\nStep 2: Numerator = 2 \u00d7 (2.0 \u00d7 10\u207b\u2075)\u00b2 \u00d7 1000 \u00d7 9.81 = 7.85 \u00d7 10\u207b\u2076 ; denominator = 9 \u00d7 1.8 \u00d7 10\u207b\u2075 = 1.62 \u00d7 10\u207b\u2074.<br>\n\nStep 3: v\u209c = 7.85 \u00d7 10\u207b\u2076 \/ 1.62 \u00d7 10\u207b\u2074 = 0.048 m\/s.<br>\n\n<strong>Answer: v\u209c \u2248 0.048 m\/s (about 4.8 cm\/s).<\/strong> The Reynolds number here is ~0.13, well below 1, which confirms Stokes&#8217; law applies \u2014 and explains why mist barely settles.\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A skydiver&#039;s terminal velocity is 53 m\/s at sea level. At a high altitude where the air density is one-quarter of the sea-level value, what would their terminal velocity be in the same posture?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong><br>\n\nStep 1: From the formula, v\u209c is proportional to 1\/\u221a\u03c1.<br>\n\nStep 2: Air density falls to \u03c1\/4, so v\u209c is multiplied by \u221a4 = 2.<br>\n\nStep 3: New v\u209c = 53 \u00d7 2 = 106 m\/s.<br>\n\n<strong>Answer: v\u209c \u2248 106 m\/s.<\/strong> This is why high-altitude jumps in thin air reach far higher speeds \u2014 the extreme version being Baumgartner&#8217;s near-vacuum record.\n\n<\/div><\/details><\/div>\n<h2>Frequently Asked Questions<\/h2>\n<details class=\"pf-faq-item\"><summary>What is terminal velocity in simple terms?<\/summary><div class=\"pf-faq-item-answer\">\n\nTerminal velocity is the fastest speed a falling object reaches in air or another fluid, when air resistance pushing up becomes as strong as gravity pulling down. At that point the forces cancel, the object stops speeding up, and it falls at a steady speed for the rest of the drop.\n\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>What is the terminal velocity of a human?<\/summary><div class=\"pf-faq-item-answer\">\n\nA human falling belly-to-earth reaches about 53 m\/s, which is roughly 120 mph or 190 km\/h. In a streamlined head-down dive, the speed rises to around 240\u2013290 km\/h (150\u2013180 mph) because the body presents a smaller area to the air. The exact figure depends on mass, posture and clothing.\n\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Do heavier objects have a higher terminal velocity?<\/summary><div class=\"pf-faq-item-answer\">\n\nFor two objects of the same shape and size, yes \u2014 terminal velocity grows with the square root of mass, so the heavier one falls faster. But shape and area matter just as much. A light, compact pebble can easily out-fall a much heavier skydiver under a parachute, because the parachute&#8217;s huge area creates enormous drag.\n\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Why doesn&#039;t terminal velocity keep increasing?<\/summary><div class=\"pf-faq-item-answer\">\n\nBecause drag grows with the square of speed. The faster you fall, the harder the air pushes back, until the upward drag exactly matches your downward weight. At that moment the net force is zero, so there is nothing left to accelerate you, and the speed stays constant.\n\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Can you reach terminal velocity on the Moon?<\/summary><div class=\"pf-faq-item-answer\">\n\nNo. Terminal velocity needs a fluid, such as air or water, to create the drag that balances gravity. The Moon has effectively no atmosphere, so there is no drag at all, and a falling object simply keeps accelerating until it lands \u2014 as Apollo 15 demonstrated when a hammer and feather fell together.\n\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>How long does it take a skydiver to reach terminal velocity?<\/summary><div class=\"pf-faq-item-answer\">\n\nA skydiver in the belly-to-earth position reaches about 99% of terminal velocity in roughly 10 to 12 seconds, after falling around 450 metres (about 1,500 feet). The approach is gradual rather than sudden \u2014 the speed rises quickly at first, then levels off as drag closes in on the weight.\n\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>Terminal velocity is the steady top speed a falling object reaches when air drag balances its weight. Learn the formula, see worked examples from skydivers to raindrops, and try an interactive lab.<\/p>\n","protected":false},"author":1,"featured_media":227,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[96,94,97,99,98,95],"class_list":["post-226","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mechanics","tag-air-resistance","tag-drag-force","tag-free-fall","tag-newtons-laws","tag-skydiving-physics","tag-terminal-velocity"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/226","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=226"}],"version-history":[{"count":1,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/226\/revisions"}],"predecessor-version":[{"id":229,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/226\/revisions\/229"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/227"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=226"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=226"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=226"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}