Free Fall & Variable Acceleration

Drop a stone down a well and count. Three seconds later you hear the splash. How deep is the water? The counting is easy; turning it into a number is not obvious, because part of those three seconds was stone falling and part was sound coming back up, and nothing in the problem tells you where the split is.

That question, and most of the ones in this unit, come down to a single fact about the world: near the ground, with air resistance out of the picture, every object falls with the same downward acceleration. Not the same speed: the same acceleration. A hammer and a feather dropped together on the Moon land together, and the astronauts of Apollo 15 filmed it to prove the point. Once you accept that the acceleration is a known constant, the graphs from the previous lesson do the rest of the work.

The one number and the three equations

Call the downward acceleration $g$. Near the Earth's surface $g = 9.8$ m/s², which means a falling object gains 9.8 m/s of downward speed every second. After one second of falling from rest it is moving at 9.8 m/s, after two seconds 19.6 m/s, after three seconds 29.4 m/s. That is the whole answer to "a wrench is dropped and falls for 3.0 s, how fast is it going".

$g$ is a positive number, always. Whether gravity speeds an object up or slows it down is decided by the sign you attach to it, not by $g$ itself. Choose a positive direction first and write it on your paper. From here on, up is positive, so the acceleration is

$a = -g = -9.8\ \mathrm{m/s^2}$

and it is $-9.8$ on the way up, at the very top, and on the way down. Gravity does not know or care which way the object happens to be moving.

Now recall the three constant-acceleration results from the previous lesson. Velocity changes at a steady rate, so it is linear in time. Position is the area under the velocity graph, which is a trapezium, so it is quadratic in time. And eliminating $t$ between those two gives a relation with no time in it at all. Written with $y$ for height, $y_0$ for the starting height and $v_0$ for the starting velocity:

$v = v_0 - gt$

$y = y_0 + v_0 t - \tfrac{1}{2} g t^2$

$v^2 = v_0^2 - 2g(y - y_0)$

Read them in words. The first says the velocity falls steadily from whatever it started at. The second says the height is where you started, plus how far you would have drifted at the old velocity, minus the sag that gravity adds. The third is the one to reach for whenever a problem mentions two heights and two speeds but never mentions time.

Three graphs that never change shape

Every vertical-motion problem here has the same three graphs behind it. The acceleration graph is a flat line at $-9.8$. The velocity graph is a straight line sloping down at that rate, crossing zero at the top of the flight. The height graph is a parabola whose peak sits directly above that crossing.

Two facts fall straight out of the velocity line. First, the object reaches its highest point when $v = 0$, which by $v = v_0 - gt$ happens at

$t_{\text{top}} = \frac{v_0}{g}$

A ball thrown up at 14.7 m/s takes $14.7/9.8 = 1.5$ s to reach the top. Second, the velocity line is symmetric about that instant: 1.5 s before the top the velocity was $+14.7$, and 1.5 s after the top it is $-14.7$. So an object passes every height twice, with the same speed both times, opposite in direction. A ball thrown up from the ground at 25 m/s takes $25/9.8 = 2.55$ s to rise, and the same again to come back, giving 5.1 s in the air.

Put $v = 0$ into the third equation and the maximum rise above the launch point is $v_0^2/2g$. For $v_0 = 20$ m/s that is $400/19.6 = 20.4$ m. Notice it goes as the square of the launch speed: throw twice as fast and you go four times as high.

Worked example 1: speed partway down

A ball is thrown straight up at 15 m/s from the top of a 50 m tower. How fast is it moving when it is 20 m above the ground?

The problem gives two heights and one speed and asks for another speed. No time is mentioned anywhere, so use the third equation. Put the origin at the ground, so $y_0 = 50$ m and $y = 20$ m, with $v_0 = +15$ m/s.

$v^2 = (15)^2 - 2(9.8)(20 - 50)$

$v^2 = 225 - 2(9.8)(-30) = 225 + 588 = 813$

$v = \sqrt{813} = 28.5\ \mathrm{m/s}$

The square root is what makes this work: $v^2$ does not care about the sign of $v$, so the answer covers both the upward and the downward pass, and here only the downward one is below the tower top. The question asked for a speed, so 28.5 m/s is the answer. Had it asked for velocity, the answer would be $-28.5$ m/s.

Also notice that the launch direction dropped out. Thrown down at 15 m/s from the same tower, the ball passes 20 m at exactly the same speed, because $v_0$ appears only as $v_0^2$.

When you cannot avoid the quadratic

Sometimes a problem insists on a time, and then you have to solve $y = y_0 + v_0 t - \tfrac12 g t^2$ for $t$, which is a quadratic. That is not a sign you have gone wrong. It is the normal state of affairs whenever the object starts and ends at different heights, because the flight is no longer symmetric.

Worked example 2: thrown up from a roof

A ball leaves a 25 m roof at 18 m/s straight up. When does it hit the ground?

Ground level is $y = 0$, and the release point is $y_0 = 25$ m with $v_0 = +18$ m/s. Set $y = 0$:

$0 = 25 + 18t - 4.9t^2$

Rearrange so the $t^2$ term is positive:

$4.9t^2 - 18t - 25 = 0$

$t = \frac{18 \pm \sqrt{(-18)^2 - 4(4.9)(-25)}}{2(4.9)} = \frac{18 \pm \sqrt{324 + 490}}{9.8}$

$t = \frac{18 \pm 28.53}{9.8}$

The two roots are $t = 4.75$ s and $t = -1.07$ s. The negative root is real mathematics, not a mistake: it is when the ball would have left the ground if the whole parabola were extended backwards. It is not the answer here, so discard it and keep $t = 4.75$ s.

The landing speed comes free: $v = 18 - 9.8(4.75) = -28.5$ m/s, so the ball lands at 28.5 m/s. Compare that with the 28.53 hiding inside the square root a moment ago. That is not a coincidence: the discriminant contains $v_0^2 + 2gh$, which is precisely $v^2$ at the ground.

Two objects, one clock

A problem with two moving objects looks twice as hard and usually is not, provided you write a position equation for each with the same $t$ and the same origin, then say in algebra what the question says in words.

Take two balls released at the same instant: one dropped from 60 m, one thrown up from the ground at 30 m/s. Where do they meet?

$y_1 = 60 - 4.9t^2, \qquad y_2 = 30t - 4.9t^2$

Meeting means $y_1 = y_2$. Subtract:

$60 - 4.9t^2 = 30t - 4.9t^2 \;\Longrightarrow\; 60 = 30t \;\Longrightarrow\; t = 2.0\ \mathrm{s}$

The $4.9t^2$ terms cancelled, and that cancellation is the real content of the problem. Both balls are pulled down identically, so relative to each other gravity does nothing at all: the gap between them closes at the constant 30 m/s the thrown ball started with. Sixty metres at 30 m/s is two seconds. Then put $t = 2.0$ into either equation:

$y = 60 - 4.9(2.0)^2 = 60 - 19.6 = 40.4\ \mathrm{m}$

The same cancellation solves problems that look unrelated. Two stones dropped from the same bridge one second apart: the moment the second is released, the first is already moving at 9.8 m/s and is 4.9 m below. After that the gap grows at a steady 9.8 m/s because both accelerate alike. Two seconds later (three seconds after the first release) the separation is $4.9 + 9.8(2.0) = 24.5$ m.

Motion in stages

Many of the harder problems are not one motion but two glued together. The rule is the same every time: solve the first stage completely, then use its final position and final velocity as the starting conditions of the second. Do not try to write one equation across the join, because the acceleration is different on each side of it.

A model rocket climbing at 4.0 m/s² for 20 s is a clean case. At burnout it has reached $v = 4.0 \times 20 = 80$ m/s and a height of $\tfrac12 (4.0)(20)^2 = 800$ m. Only now does free fall begin, from 800 m with $v_0 = +80$ m/s, adding $80^2/(2 \times 9.8) = 326.5$ m. Maximum height: 1126.5 m. A reader who forgets the 800 m, or who uses $g$ during the burn, will be out by hundreds of metres.

A bouncing ball is the same idea. Dropped from 5.0 m it hits at $\sqrt{2(9.8)(5.0)} = 9.90$ m/s after $\sqrt{2(5.0)/9.8} = 1.01$ s. If it rebounds with 80% of that, it leaves the floor at 7.92 m/s, and a throw up at 7.92 m/s returns to the floor after $2(7.92)/9.8 = 1.62$ s. Second impact at $1.01 + 1.62 = 2.63$ s.

Worked example 3: the depth of the well

Back to the stone and the splash, 3.00 s after release, with sound at 340 m/s.

Two stages again, but this one has a twist: you are told the total time and neither piece of it. Name the unknown depth $d$ and write each stage in terms of it. The fall takes $t_1$ with $d = \tfrac12 g t_1^2$, so $t_1 = \sqrt{2d/g}$. The sound comes back up at constant speed, taking $t_2 = d/340$. Their sum is what you heard:

$\sqrt{\frac{2d}{9.8}} + \frac{d}{340} = 3.00$

That is awkward in $d$, so use $t_1$ as the unknown instead, since $d = 4.9t_1^2$:

$t_1 + \frac{4.9t_1^2}{340} = 3.00$

$0.014412\,t_1^2 + t_1 - 3.00 = 0$

$t_1 = \frac{-1 + \sqrt{1 + 4(0.014412)(3.00)}}{2(0.014412)} = \frac{-1 + \sqrt{1.17294}}{0.028824}$

$t_1 = \frac{0.08302}{0.028824} = 2.880\ \mathrm{s}$

$d = 4.9(2.880)^2 = 40.7\ \mathrm{m}$

Check it against the crude answer. Ignoring the sound entirely gives $4.9(3.00)^2 = 44.1$ m, and the true depth is a little less than that, as it must be, since some of the three seconds was spent on the return trip. The sound leg took $40.7/340 = 0.12$ s, which is small but not nothing.

When the acceleration refuses to stay constant

Everything so far assumed no air. Real falls do not keep accelerating forever; a skydiver levels off near 55 m/s and a raindrop near 9 m/s. Drag pushes back against the motion and grows with speed, so the net downward force shrinks as the object speeds up. Write the slow-speed case, where drag is proportional to speed with some constant $b$:

$m\frac{dv}{dt} = mg - bv$

At $v = 0$ the acceleration is the full $g$. As $v$ climbs, the right-hand side shrinks. It reaches zero at the terminal speed, where the object stops accelerating and falls forever at a fixed rate:

$v_t = \frac{mg}{b}$

You do not need calculus to find $v_t$: set the net force to zero and solve. Getting the whole history $v(t)$ does need calculus, and the result you will be given is

$v(t) = v_t\left(1 - e^{-gt/v_t}\right)$

Check the two ends. At $t = 0$ the bracket is $1 - 1 = 0$, as it should be. For small $t$, $e^{-x} \approx 1 - x$, so $v \approx v_t (gt/v_t) = gt$, the no-drag answer: drag barely matters at the start. For large $t$ the exponential dies and $v \to v_t$ from below. The object never overshoots.

With $v_t = 6.0$ m/s, the speed at $t = 1.20$ s is

$v = 6.0\left(1 - e^{-(9.8)(1.20)/6.0}\right) = 6.0\left(1 - e^{-1.96}\right)$

$e^{-1.96} = 0.1409, \qquad v = 6.0(0.8591) = 5.15\ \mathrm{m/s}$

Constant acceleration would have predicted $9.8 \times 1.20 = 11.8$ m/s, nearly twice as much. For fast, blunt objects the drag goes as $v^2$ instead, terminal speed becomes $v_t = \sqrt{mg/c}$, and the useful form is in terms of distance fallen $y$ rather than time:

$v^2 = v_t^2\left(1 - e^{-2gy/v_t^2}\right)$

Reaching 40 m/s with $v_t = 50$ m/s needs $1600 = 2500(1 - e^{-2gy/2500})$, so $e^{-2gy/2500} = 0.36$, giving $2(9.8)y/2500 = \ln(1/0.36) = 1.0217$ and $y = 130$ m. Both formulas have the same skeleton: a terminal value, and an exponential that measures how much of the way there you are.

Recognising a free-fall problem

The tell is that gravity is the only thing acting and the motion is along one vertical line. Words like dropped, released, thrown straight up, falls freely, neglect air resistance all point here. Once you are sure, three questions pick the equation for you.

Does the problem mention time, or ask for it? If yes, you need $v = v_0 - gt$ or the quadratic $y = y_0 + v_0 t - \tfrac12 gt^2$. If time is never mentioned in either the given data or the question, use $v^2 = v_0^2 - 2g(y-y_0)$ and save yourself a step.

Is there more than one moving object, or more than one phase of motion? Two objects means two position equations sharing one clock, then set them equal or subtract them, and expect the $t^2$ terms to cancel. Two phases, like a rocket burn then a coast, or a fall then a bounce, means solving the first stage fully and feeding its end into the start of the second.

Does the problem mention air resistance, terminal speed, or give you an exponential? Then constant acceleration is dead and you are plugging into the drag formulas above. The giveaway is any phrase about a speed that stops increasing.

One last habit worth building: before solving, estimate. A drop of a few metres takes about a second and ends near 10 m/s; a ten-storey fall takes about two and a half seconds. If your algebra returns 300 m/s for a stone off a bridge, the sign of $g$ is probably wrong somewhere.

Every motion so far has been along a single vertical line. Next you keep the same $-g$ but give the object a sideways velocity too, and find that the horizontal and vertical halves of the motion never talk to each other. That is two-dimensional and projectile motion.