2.1 - Uniform acceleration and free fall

Equations for constant acceleration

If a ball is released from four times the height, does it take four times as long to land? It is tempting to scale the time with the distance. But the ball gains speed as it falls, so later seconds cover more distance than earlier seconds. A model with constant acceleration lets you predict the correct scaling.

First separate the model from the world. We are describing motion along one straight line with constant acceleration. The initial velocity is $u$, final velocity $v$, displacement $s$, elapsed time $t$ and acceleration $a$. Choose a positive direction once. Velocities have unit $\mathrm{m\,s^{-1}}$, displacement metres, time seconds and acceleration $\mathrm{m\,s^{-2}}$.

From the definition of acceleration,

A constant acceleration gives a straight velocity–time line. Its area is a rectangle plus a triangle, or one trapezium. Both represent displacement.

[DIAGRAM: asset_name: The area that generates the motion equations; asset_slug: motion-constant-acceleration-area; caption: Rectangle plus triangle gives the displacement for a straight velocity-time line; description: A velocity-time line rises from u equal to 2 metres per second at zero time to v equal to 10 metres per second at 4 seconds. A blue rectangle of height 2 and width 4 lies below a pale clay triangle of height 8. The total area is 24 metres, matching the trapezium expression half times u plus v times t.]

The rectangle has area $ut$; the triangle has area $\tfrac12(v-u)t$. Therefore

Eliminate time instead: multiply $s=\tfrac12(u+v)t$ by $2a$, then use $at=v-u$.

These derivations are why you must not apply the equations to a changing acceleration by habit. The straight graph, and hence its simple triangular area, is part of the reasoning.

The two supplied equations are not the entire toolkit. From their derivations you should also know $v=u+at$, $s=\tfrac12(u+v)t$, and, by replacing $u$ with $v-at$, $s=vt-\tfrac12at^2$. These three forms are not printed on the general data sheet: learn them. The equations describe signed displacement even if the object reverses direction; total distance then needs the journey split at the reversal.

Signed motion and free fall

The easiest equation is usually the one that omits a quantity you do not know and do not need. List the known signed values, identify the target and check that acceleration can be treated as constant before substituting.

In free fall, gravity is the only force acting. Near Earth's surface, and over modest changes in height, the acceleration can be treated as constant and downwards. Neglect air resistance. The data sheet gives its magnitude as $g=9.81\,\mathrm{m\,s^{-2}}$. If upwards is positive, $a=-g$, even while the object is rising. “Released from rest” means $u=0$; “thrown” does not.

For a release from rest through a downward height $h$, choose downwards positive. The supplied equation becomes $h=\tfrac12gt^2$, so $t=\sqrt{2h/g}$. Increasing $h$ by a factor of four increases the time by a factor of two. For example, ideal falls of $0.80\,\mathrm{m}$ and $3.2\,\mathrm{m}$ take about $0.404\,\mathrm{s}$ and $0.808\,\mathrm{s}$. The larger height is covered at a greater average speed.

Some flight-time problems give a quadratic equation. Recall the mathematical tool rather than repeatedly trying numbers: if $At^2+Bt+C=0$, then

The quadratic formula is not on the physics data sheet: learn it. Here capital $A$, $B$ and $C$ are coefficients, not accelerations; their units must be consistent with the equation. Interpret both roots in the physical time interval.

Choose signs before using the supplied relationship.

Measuring the acceleration of free fall

Timing a short fall by hand can give reaction-time uncertainty that is large compared with the fall time. An electronic release and impact switch make the start and finish events part of the apparatus instead of part of your reaction.

Clamp an electromagnet above an impact switch connected to an electronic timer. A steel ball is held at rest. Releasing it starts the timer; its first contact with the switch stops the timer. Measure $h$ from the bottom of the ball at release to the contact surface of the switch. This is the distance the ball travels before the stopping event, rather than the distance from the magnet or the ball's centre to the base.

[DIAGRAM: asset_name: An electronic free-fall measurement; asset_slug: motion-free-fall-timer; caption: Measure the ball's travel from its lower surface at release to first contact; description: A clamped electromagnet holds a steel ball above a horizontal impact switch and catching tray. A vertical dimension h starts at the bottom of the held ball and ends at the top contact surface. Dashed electrical leads connect release and impact events to a timer. A labelled stand supports the magnet.]

Vary the height over a useful range. At each height, repeat the release, check for unusual readings and calculate a mean time. Keep the same ball and release mechanism, and do not push the ball down. A low, clear drop zone with a catching tray reduces the risk of a ball striking feet or rolling away. Secure the stand so it cannot topple.

Repeats reduce the effect of random timing variation; they do not remove a systematic delay between physical release and timer start. A constant height offset may give a non-zero intercept. A release delay can distort the height–time-squared relationship more subtly. Check how the timer is triggered and measure the actual event-to-event travel distance. Increasing the fall distance reduces relative height uncertainty, within safe limits, but a very long fall can make air resistance more significant.

Quick Recap

• Constant-acceleration equations follow from the definition of acceleration and the area under a straight velocity–time line.
• Choose one positive direction; free-fall acceleration stays downwards even while an object rises.
• Use the quadratic formula when necessary and keep only roots relevant to the physical interval.
• Electronic release/impact timing, repeated readings and a graph of $h$ against $t^2$ give a practical measurement of $g$.

For an ideal release from rest, four times the height gives twice the fall time, because $h$ is proportional to $t^2$.