Vertical motion under gravity

Why an object moving freely under gravity is just a constant-acceleration problem with a = ±g, how to choose a positive direction and translate "thrown up", "returns", "hits the ground" into suvat values, and why the up-and-down trip makes distance and displacement differ.

Mechanics · Kinematics

Vertical motion under gravity

Why it works

"Moving freely under gravity" is a modelling phrase with a precise meaning: the only force acting is gravity. That makes the acceleration constant — its size is

g \approx 9.8\ \text{m s}^{-2}

(some questions tell you to use

10\ \text{m s}^{-2}

), always directed downwards. Constant acceleration means the whole toolkit you already have applies: vertical motion under gravity is just suvat with $a = \pm g$ . Nothing new — except discipline with signs.

Everything hinges on one choice: which way is positive. Pick a direction, then apply it to every quantity for the whole question.

If you take up as positive, gravity acts the other way, so $a = -g$ . A

downward velocity or displacement is then negative.

If you take down as positive, $a = +g$ .

A frequent trap: gravity does not switch sign at the top. On the way up it slows the object; at the top the velocity is momentarily zero; on the way down it speeds the object up — but it is the same downward acceleration

g

the whole time. The object doesn't hang at the top because the acceleration there is still

g

, not zero.

Translating the words into suvat values is where most marks are won or lost (taking up as positive throughout):

"dropped" / "released from rest" $\Rightarrow u = 0$ ;
"thrown/projected up at $u$ " $\Rightarrow$ initial velocity $= +u$ ;
"at its maximum height" $\Rightarrow v = 0$ ;
"returns to its starting point" $\Rightarrow s = 0$ ;
"hits the ground", from a height $h$ above it, $\Rightarrow s = -h$ ;
"arrives moving downwards at speed $w$ " $\Rightarrow v = -w$ .

That last one matters: a question often gives you the speed on arrival, and you must attach the sign yourself from the direction of travel.The velocity–time graph above (taking up as positive,

u = 20

g = 10

) shows it all at once: a straight line of gradient $-g$ — constant downward acceleration throughout, including at the top where it merely crosses zero. The area above the axis is the rise; the area below is the fall.

Distance vs displacement on the round trip. Because the object goes up and then comes back down, once it passes its start the suvat

s

(which is net displacement) is not the total distance travelled. To get distance, split the journey at the highest point (where

v = 0

): find the rise, find the fall, and add their sizes — exactly the split-at-

v=0

idea from [[kinematics.integrating-motion]], here forced by gravity.

The model and its refinements. "Freely under gravity" assumes no air resistance and treats the object as a particle. So besides air resistance, you can refine the model by accounting for the object's size and shape (it isn't really a point), wind, or spin. And if air resistance were included, it opposes the motion: the object would not climb as high, and to still arrive at a given speed it would have to be thrown faster — so predictions about speeds and required launch velocities shift in a definite direction.