Motion in a Straight Line · Lesson 6 of 10
Free Fall and Vertical Motion
“Apply constant acceleration to falling and thrown objects, including sign conventions and Galileo’s odd-number law.”
• Model free fall with constant downward acceleration. • Derive rise time, maximum height and return time. • Solve a throw from a building by two methods. • Explain Galileo’s odd-number law and interpret vertical-motion graphs.
Free fall includes upward motion
An object is in free fall when gravity is the only significant force acting on it. It need not already be moving downward. A ball thrown upward is in free fall after it leaves the hand, provided air resistance is negligible. Near Earth’s surface, over heights small compared with Earth’s radius, take g approximately constant at 9.8 m/s². Some calculations explicitly use 10 m/s² for convenience; keep the chosen value throughout that calculation.
Choose upward as positive. Then acceleration is a = −g during ascent, at the highest point, and during descent. Gravity does not reverse when the ball reverses. If downward is chosen positive instead, acceleration is +g. The physical prediction must remain the same under either convention. Mass does not enter these idealised equations, so two released objects at the same location fall together when air resistance is negligible.
Dropping from rest
Released from rest means u = 0, not that acceleration is zero. With origin at the release point and upward positive, y = −gt²/2 and v = −gt. The position becomes more negative and speed grows as gt. Downward distance d is positive and equals gt²/2. Thus the acceleration-time graph is a horizontal line at −g, velocity-time is a straight line with slope −g, and position-time is a downward-opening parabola. The graph of distance fallen is instead an upward-opening parabola.
Problem
An object is released from rest and falls for 2 s with g = 9.8 m/s². Find velocity, displacement and distance.
- 1.With upward positive, v = −9.8 × 2 = −19.6 m/s.
- 2.Displacement is −½ × 9.8 × 2² = −19.6 m.
- 3.Distance fallen is +19.6 m. The matching numerical values of velocity and displacement here are coincidental; their units and meanings differ.
Throwing upward from a level surface
For initial upward speed u > 0, the ball slows because positive velocity and negative acceleration have opposite signs. At the top, set v = 0 in v = u − gt to obtain rise time u/g. Set v = 0 in v² = u² − 2gH to obtain the height gained H = u²/(2g). At this instant, velocity and speed are zero but acceleration remains −g. Afterward, both v and a are negative, so the ball speeds up downward.
To derive the return time, set y − y₀ = 0: 0 = ut − gt²/2 = t(u − gt/2). The root t = 0 describes launch; the later root is 2u/g. Substitution into v = u − gt gives −u at return. At any height reached on both ascent and descent, speed is the same but velocity has the opposite sign. This symmetry fails if drag matters or if you compare unequal heights.
Problem
Take g = 9.8 m/s². Find the greatest height above release and time to return.
- 1.Rise time is 29.4/9.8 = 3.0 s.
- 2.Height gained is 29.4²/(2 × 9.8) = 44.1 m.
- 3.Return time is 6.0 s and return velocity is −29.4 m/s with upward positive.
- 4.Acceleration is −9.8 m/s² throughout the flight, including the top.
Moving the origin to the highest point
Now put y = 0 and t = 0 at the top and choose downward positive. Every other point on the ascent and descent lies below the top, so its position y is positive. During ascent, velocity is negative; during descent, it is positive. Acceleration is +g in both phases. Times before the top are negative under this clock choice. The equations y = gt²/2 and v = gt show all these signs consistently. Negative time simply means “before the chosen clock zero”.
Throwing from a building: two solution routes
Problem
A ball leaves a roof 25 m above the ground with upward velocity 20 m/s. Take g = 10 m/s². Find maximum height and time to hit the ground.
- 1.Choose ground as y = 0 and upward positive. The extra rise is u²/(2g) = 20 m, so the maximum height above ground is 45 m.
- 2.Method A: rise time = 20/10 = 2 s. From the top it falls 45 m from rest, so 45 = ½ × 10 × t_down² and t_down = 3 s. Total flight time = 5 s.
- 3.Method B: use the entire flight. At impact, 0 = 25 + 20t − 5t². Rearranging gives (t − 5)(t + 1) = 0. The physical root after launch is t = 5 s; t = −1 s belongs to the backward extension of the mathematical trajectory.
- 4.Impact velocity is 20 − 10 × 5 = −30 m/s. Total distance is 20 + 45 = 65 m, while displacement is 0 − 25 = −25 m.
Galileo’s odd-number law
For a body released from rest under constant acceleration, distance from release grows as t². Compare equal successive durations τ. After τ, 2τ, 3τ and 4τ, the distances from the start are proportional to 1, 4, 9 and 16. The extra distances in the individual intervals are therefore proportional to 1, 3, 5 and 7. These are successive-interval distances, not distances measured from the starting point.
| Time since release | Total distance divided by ½gτ² | Distance in latest interval divided by ½gτ² |
|---|---|---|
| τ | 1 | 1 |
| 2τ | 4 | 3 |
| 3τ | 9 | 5 |
| 4τ | 16 | 7 |
| 5τ | 25 | 9 |
For nonzero initial velocity, the displacement in the nth interval includes an extra uτ term. The pure odd-number ratio then does not generally hold. If an interval contains a direction reversal, distinguish its displacement from its distance before applying an interval formula. These conditions are part of the law, not optional details.
Quiz
At the highest point of a freely thrown ball, neglecting air resistance, which statement is correct?
Key Takeaways
• Gravity acts downward through the entire free-flight phase. • At the top, velocity is zero but acceleration is not. • Use 2u/g only for a return to the launch height. • Separate height above ground, height gained, displacement and distance. • Galileo’s odd-number law concerns equal successive intervals after release from rest.