Free Fall & Vertical Motion

Once a body is in flight and air resistance is negligible, the only force on it is its weight, W = mg. Dividing by the mass gives the acceleration, a = W/m = g, and the mass has cancelled: that is why the hammer and the feather David Scott dropped on the airless Moon in 1971 landed together. The weight points toward Earth's centre however the body is moving, so the acceleration is 9.81 m s⁻² downward on the way up, at the top and on the way down. Its size depends on place, not on motion: about 9.78 m s⁻² at the equator, 9.83 m s⁻² at the poles, and 1.62 m s⁻² on the Moon, where a 2.0 m drop takes 1.57 s instead of Earth's 0.64 s. The model covers only the stretch between the instant contact ends and the instant something touches the body again. While a hand is still pushing, the acceleration is whatever the hand makes it, which is why every free-fall clock starts at release.

g is a magnitude and is always positive; the acceleration is a signed component whose sign depends on the axis you declare. With up positive, a = −g, an upward launch has v₀ > 0, a downward throw has v₀ < 0, and a landing below the start has y − y₀ < 0. With down positive every one of those signs flips and a = +g. Neither choice is more correct, but mixing them inside one problem is the commonest source of wrong answers. Take a stone thrown down at 4.0 m s⁻¹ from a bridge 20.0 m above the water. Up positive: 0 = 20.0 − 4.0t − 4.905t², so t = (−4.0 + √408.4)/9.81 = 1.65 s and v = −4.0 − 9.81 × 1.65 = −20.2 m s⁻¹. Down positive: 20.0 = 4.0t + 4.905t² gives the same 1.65 s and v = +20.2 m s⁻¹. The physics, 20.2 m s⁻¹ downward after 1.65 s, does not care which convention produced it.

The free-fall clock starts when contact ends, and v₀ is the body's velocity at that instant, measured from the ground. Letting go adds nothing to the velocity the body already shares with whatever carried it; throwing adds the velocity of the throw relative to the carrier. So a sandbag let go from a balloon climbing at 2.5 m s⁻¹ does not start falling at once. It leaves at v₀ = +2.5 m s⁻¹, keeps rising for 2.5/9.81 = 0.25 s, gains another 2.5²/(2 × 9.81) = 0.32 m, and only then descends. A ball tossed upward at 6.0 m s⁻¹ relative to a lift that is descending at 2.0 m s⁻¹ leaves at v₀ = −2.0 + 6.0 = +4.0 m s⁻¹, while a coin simply let go in the same lift leaves at −2.0 m s⁻¹. “Dropped” means released with the carrier's velocity, and it means “from rest” only when the carrier is at rest.

At the top of a vertical flight the velocity, not the acceleration, is zero: v = v₀ − gt = 0 gives tₜₒₚ = v₀/g, and the height gained is v₀²/(2g). Because the acceleration is identical either side of that instant, the flight is symmetric in time about tₜₒₚ, and every height below the apex is passed twice at the same speed. Launch at 12.0 m s⁻¹ from the ground and ask when the body is 5.00 m up. Solving 5.00 = 12.0t − 4.905t² gives t = 0.533 s and 1.914 s. Their average is tₜₒₚ = 1.223 s, and at both passes the speed is √(12.0² − 2 × 9.81 × 5.00) = 6.78 m s⁻¹, upward then downward. Both roots are real events. That changes when the target lies below the launch point: thrown up at 12.0 m s⁻¹ from a 5.00 m balcony, the ground equation gives t = 2.81 s and −0.36 s. The negative root is when the same parabola, extended backwards, crossed the ground before the throw, so it is no part of this flight.

Plotted against time, every vertical flight has the same signature. Completing the square turns y = y₀ + v₀t − ½gt² into y = yₜₒₚ − ½g(t − tₜₒₚ)², so every launch traces one and the same parabola, shifted until its vertex sits on the apex. The v–t graph is a straight line of slope −9.81 m s⁻² that crosses zero at tₜₒₚ and carries straight on: it does not bend, flatten or bounce at the top. The a–t graph is a horizontal line at −9.81 m s⁻² from release to landing that never touches zero. A V-shaped graph with its point on the time axis is the speed |v|, not the velocity, and drawing it for v is the graph form of believing the ball stops accelerating at the top. Launch faster and the v–t line slides up parallel to itself; launch from higher and the same line simply runs longer before the ground ends it.

Three approximations hold the model up, and each fails at a known scale. The field is taken as uniform, but g weakens with distance from Earth's centre as g₀R²/(R + h)² ≈ g₀(1 − 2h/R): only 0.31% weaker 10 km up, and still 8.7 m s⁻² at the 400 km altitude of the International Space Station, whose crew are in free fall, not beyond gravity. Air resistance is taken as negligible, and that fails far sooner. For drag growing as v², the drag-to-weight ratio is (v/vₜ)², where vₜ is the terminal speed, so drag reaches 10% of the weight at v = 0.32 vₜ. A skydiver with vₜ ≈ 55 m s⁻¹ gets there after about 1.8 s of fall; a table-tennis ball with vₜ ≈ 9 m s⁻¹ after falling only 0.41 m. The ground is taken as a non-rotating frame, yet Earth's rotation lands a 100 m drop at 45° latitude about 1.6 cm east of the plumb line. For dense objects thrown by hand, drag is the approximation to test first.