Physics
Free Fall: Formulas and 7 Solved Exercises
In free fall v = g · t and h = ½ · g · t², with g = 9.8 m/s². Here are the three formulas, the metres-per-second table, typical errors and 7 solved exercises.
Recommended for: Grade 9 · Grade 10
A body is in free fall when gravity is the only force acting on it and air resistance can be ignored. The acceleration is the same for every body, g = 9.8 m/s² pointing downwards, and the formulas to know are v = g · t, h = ½ · g · t² and v² = 2 · g · h.
The three formulas and when to use them
| Formula | Use it when | It gives you |
|---|---|---|
| v = g · t | you know the time of fall | the speed at that instant |
| h = ½ · g · t² | you know the time of fall | the distance fallen |
| v² = 2 · g · h | time is neither given nor asked for | the speed from the height |
These three hold for a body released from rest. With an initial speed v₀ they become v = v₀ + g · t, h = v₀ · t + ½ · g · t² and v² = v₀² + 2 · g · h: the same formulas as uniformly accelerated motion, with a replaced by g.
A fourth, very handy formula follows from the second one and gives the time of fall: t = √(2 · h / g). You need it whenever the problem hands you the height and asks how long the fall lasts.
Table: how far a body falls each second
| Time | Distance fallen h | Speed v | Distance in that second |
|---|---|---|---|
| 1 s | 4.9 m | 9.8 m/s | 4.9 m |
| 2 s | 19.6 m | 19.6 m/s | 14.7 m |
| 3 s | 44.1 m | 29.4 m/s | 24.5 m |
| 4 s | 78.4 m | 39.2 m/s | 34.3 m |
| 5 s | 122.5 m | 49.0 m/s | 44.1 m |
The right-hand column is what gives the motion away: unlike uniform motion, where the body covers the same distance every second, here each second adds 9.8 m to the one before. In the first 5 s a body falls 122.5 m, and more than a third of that distance happens in the last second alone.
Why mass makes no difference
This is the point that costs the most marks on theory questions. The weight of a body is W = m · g, so a body twice as heavy is pulled down twice as hard: but it also has twice as much mass to set in motion. In Newton’s second law m · a = m · g the mass cancels, leaving a = g, identical for everything.
In everyday life a feather and a coin do not land together, but the culprit is the air. In a tube with the air pumped out the feather falls exactly like the coin, and the experiment has even been repeated on the Moon, where there is no atmosphere at all.
Throwing upwards is still free fall
As long as gravity is the only force, the motion is free fall even while the body is rising. It helps to take upwards as positive and write the acceleration as −g. Two practical consequences:
- at the highest point the speed is zero, but the acceleration is still 9.8 m/s²;
- rise and fall between the same two heights take the same time and happen at the same speed in magnitude.
When the body starts at some height and lands lower down, setting y = h + v₀ · t − ½ · g · t² equal to zero gives a quadratic equation. That is exactly the situation in the last exercise, and it is solved with the formula covered in quadratic equations.
The value of g elsewhere
| Body | g (m/s²) | Time to fall 20 m |
|---|---|---|
| Earth | 9.81 | 2.02 s |
| Moon | 1.62 | 4.97 s |
| Mars | 3.72 | 3.28 s |
| Jupiter | 24.8 | 1.27 s |
The third column comes from t = √(2 · h / g) and is a good way to check that you have understood the formula: with g six times smaller the time does not become six times longer but only about two and a half times, because g sits under a square root.
The most common mistakes
- Dividing the height by the time to get the final speed: that gives the average speed, which in free fall is exactly half the final one.
- Forgetting the ½ in h = ½ · g · t²: the mistake that doubles every answer.
- Putting mass into the formulas: it never appears, not even when the question gives it to you (it is often there on purpose as a red herring).
- Using v = g · t when there is an initial speed: if the body is thrown rather than released, you need v = v₀ + g · t.
- Mixing up units: convert km/h to m/s by dividing by 3.6 before using any formula.
- Confusing total height with the distance in the last second: that has to be computed as the difference h(t) − h(t − 1).
A four-step method
First, list the data with units and convert whatever needs converting. Second, choose the positive direction and write the sign of g consistently (usually positive if you pick downwards, negative if you pick upwards). Third, pick the formula that does not contain the quantity you are missing. Fourth, check the answer with a different formula, as is done in every exercise below.
Solved exercises
1. You drop a stone from a bridge and it hits the water after 2 s. How fast is it going on impact, and how high is the bridge above the water? base
Show solution
- The stone is released, so v₀ = 0 and free fall applies with g = 9.8 m/s².
- Speed: v = g · t = 9.8 · 2 = 19.6 m/s.
- Height: h = ½ · g · t² = ½ · 9.8 · 2² = ½ · 9.8 · 4 = 19.6 m.
- The two numbers match only because t = 2 s: they have different units (m/s and m) and must not be confused.
Answer: v = 19.6 m/s ; h = 19.6 m
2. An object is released from rest at a height of 78.4 m. How long does it take to reach the ground, and at what speed does it land? base
Show solution
- Rearrange h = ½ · g · t² for the time: t = √(2 · h / g).
- t = √(2 · 78.4 / 9.8) = √(156.8 / 9.8) = √16 = 4 s.
- Speed: v = g · t = 9.8 · 4 = 39.2 m/s.
- Check with the time-free formula: v² = 2 · g · h = 2 · 9.8 · 78.4 = 1536.64 and √1536.64 = 39.2. It matches.
Answer: t = 4 s ; v = 39.2 m/s
3. From the same window, 20 m above the ground, you release a 2 kg lead sphere and a 0.2 kg plastic ball at the same instant. Which one lands first, and after how long? base
Show solution
- Free fall ignores air resistance, so mass appears in none of the formulas: both have the same acceleration g.
- t = √(2 · h / g) = √(2 · 20 / 9.8) = √(40 / 9.8) = √4.0816… ≈ 2.02 s.
- Landing speed: v = g · t = 9.8 · 2.02 ≈ 19.8 m/s, the same for both.
- In reality the plastic ball is slowed more by the air and lands slightly later, but that is friction at work, not weight.
Answer: They land together, t ≈ 2.02 s (v ≈ 19.8 m/s)
4. From a balcony 40 m up you throw a ball downwards with an initial speed of 5 m/s. What is its speed on impact and how long does the fall last? intermedio
Show solution
- Here v₀ ≠ 0, so use the full formulas with downwards taken as positive.
- The time is unknown, so start from v² = v₀² + 2 · g · h = 5² + 2 · 9.8 · 40 = 25 + 784 = 809.
- v = √809 ≈ 28.44 m/s.
- Time: from v = v₀ + g · t you get t = (28.44 − 5) / 9.8 = 23.44 / 9.8 ≈ 2.39 s.
- Check: h = v₀ · t + ½ · g · t² = 5 · 2.39 + 4.9 · 2.39² ≈ 11.96 + 28.04 = 40.0 m. It matches.
Answer: v ≈ 28.4 m/s ; t ≈ 2.39 s
5. You throw a ball straight up with an initial speed of 19.6 m/s. What maximum height does it reach, and after how long does it land back in your hand? intermedio
Show solution
- Take upwards as positive: the acceleration is −g, that is −9.8 m/s².
- At the highest point the speed is zero: 0 = v₀ − g · t, so t_up = 19.6 / 9.8 = 2 s.
- Maximum height: h = v₀²/(2 · g) = 19.6² / (2 · 9.8) = 384.16 / 19.6 = 19.6 m.
- The way down is a free fall from 19.6 m and takes exactly as long as the way up, so t_total = 2 + 2 = 4 s.
- Check on the descent: t = √(2 · 19.6 / 9.8) = √4 = 2 s. It matches.
Answer: h_max = 19.6 m ; t_total = 4 s
6. A body falls from rest through 122.5 m. How many metres does it cover during the last second of the fall? avanzato
Show solution
- First the total time: t = √(2 · 122.5 / 9.8) = √25 = 5 s.
- The last second runs from t = 4 s to t = 5 s, so work out the distance covered in the first 4 s.
- h(4 s) = ½ · 9.8 · 4² = 4.9 · 16 = 78.4 m.
- Distance in the last second = 122.5 − 78.4 = 44.1 m.
- Note the error to avoid: it is not 122.5 / 5 = 24.5 m. In free fall the per-second distances grow as 4.9 − 14.7 − 24.5 − 34.3 − 44.1 m.
Answer: 44.1 m (more than a third of the whole fall)
7. From the top of a 25 m tower you throw a stone upwards at 15 m/s. How long before it reaches the ground, and at what speed? avanzato
Show solution
- Upwards positive, origin at the foot of the tower: y = 25 + 15 · t − 4.9 · t².
- The stone is on the ground when y = 0, that is 4.9 · t² − 15 · t − 25 = 0.
- Solve the quadratic: Δ = 15² + 4 · 4.9 · 25 = 225 + 490 = 715 and √715 ≈ 26.74.
- t = (15 + 26.74) / (2 · 4.9) = 41.74 / 9.8 ≈ 4.26 s (discard the negative root, which has no physical meaning).
- Speed: v = 15 − 9.8 · 4.26 ≈ −26.7 m/s, that is 26.7 m/s directed downwards.
- Independent check with v² = v₀² + 2 · g · h = 15² + 2 · 9.8 · 25 = 225 + 490 = 715 and √715 ≈ 26.74 m/s. It matches.
Answer: t ≈ 4.26 s ; v ≈ 26.7 m/s downwards
FAQ
Why does mass not affect free fall?
Because weight is proportional to mass (W = m · g), but so is the inertia you have to overcome. In Newton's second law, m · a = m · g, the mass cancels and you are left with a = g: every body falls with the same acceleration. In real life a feather lands after a stone only because the air slows it down, not because it is lighter.
What is the value of g, and can I use 10 m/s²?
The standard value is g ≈ 9.81 m/s², and problems almost always use 9.8 m/s². Using 10 m/s² is fine only if the question says so or you want a quick estimate: it introduces an error of about 2%, which on a test asking for two significant figures can make a correct method look wrong.
What is the difference between free fall and uniformly accelerated motion?
There is no difference in the formulas: free fall is the special case of uniformly accelerated motion in which the acceleration is gravity, equal to 9.8 m/s² and always pointing downwards. The only change is that you do not have to work out a, because you already know it.
How do you find the distance covered in the last second of a fall?
Work out the total distance and the distance up to one second earlier, then subtract: h(t) − h(t − 1). Do not divide the height by the number of seconds: in free fall the per-second distances grow as 4.9 − 14.7 − 24.5 − 34.3 m and so on.
Is a body thrown upwards still in free fall?
Yes, if you ignore the air. From the moment it leaves your hand the only acceleration is g pointing downwards, even while it is rising: that is why it slows down. At the highest point the speed is zero but the acceleration is still 9.8 m/s², which is exactly why the body does not stay up there.