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The **free fall equation** describes the relationship between the distance traveled (h), time elapsed (t), and acceleration due to gravity (g) for a falling object.

Free fall equations in terms of time:

- Distance: h = ½ × g t
^{2} - Velocity: v = g t

The distance traveled by the object during free fall can be calculated using the equation h = ½ × g t^{2}. Additionally, the velocity of the object can be determined using the equation v = g t.

Free fall equations in terms of distance:

- Time: t = √2 h/g
- Velocity: v = √2 g h

If the distance traveled during free fall is known, the time it takes for the object to fall can be calculated using the equation t = √2 h/g. The velocity of the object can also be determined using the equation v = √2 g h.

Free fall equations in terms of velocity:

- Time: t = v/g
- Distance: h = v
^{2}/2 g

If the velocity of the object during free fall is known, the time it takes for the object to fall can be calculated using the equation t = v/g. Similarly, the distance traveled by the object can be determined using the equation h = v^{2}/2 g.

## Practice problems

### Problem #1

A feather is released from the top of a building, free-falling for a time of 20 seconds. Calculate the distance traveled by the feather during this time and determine its velocity. Assume a gravitational acceleration of g = 9.81 m/s^{2}.

**Solution**

Given data:

- Time taken by a feather, t = 20 s
- Distance traveled by a feather, h = ?
- Velocity of a feather, v = ?
- Gravitational acceleration, g = 9.81 m/s
^{2}

Applying the formula of free fall distance, in terms of time:

- h = ½ × g t
^{2} - h = ½ × 9.81 × (20)
^{2} - h = ½ × 9.81 × 400
- h = 1962 m

Therefore, the distance traveled by a feather is **1962 m**.

Applying the formula of free fall velocity, in terms of time:

- v = g t
- v = 9.81 × 20
- v = 196.2 m/s

Therefore, the velocity of a feather is **196.2 m/s**.

### Problem #2

A ball is dropped from the top of a tower and falls a distance of 600 meters. Determine the time it takes for the ball to reach the ground and calculate its velocity upon reaching the ground. Use a gravitational acceleration of g = 9.81 m/s^{2}.

**Solution**

Given data:

- Distance traveled by a ball, h = 600 m
- Time taken by a ball to reach the ground, t = ?
- Velocity of a ball, v = ?
- Gravitational acceleration, g = 9.81 m/s
^{2}

Applying the formula of free fall time, in terms of distance:

- t = √2 h/g
- t = √(2 × 600)/9.81
- t = √1200/9.81
- t = √122.3241
- t = 11.06 s

Therefore, the time taken by a ball to reach the ground is **11.06 s**.

Applying the formula of free fall velocity, in terms of distance:

- v = √2 g h
- v = √2 × 9.81 × 600
- v = √11772
- v = 108.49 m/s

Therefore, the velocity of a ball is **108.49 m/s**.

### Problem #3

A small piece of paper is dropped from the 10^{th} floor of a building and falls with an initial velocity of 26 m/s. Determine the time it takes for the paper to reach the ground and calculate the distance traveled by the paper. Use a gravitational acceleration of g = 9.81 m/s^{2}.

**Solution**

Given data:

- Velocity of a paper, v = 26 m/s
- Time taken by a paper to reach the ground, t = ?
- Distance traveled by a paper, h = ?
- Gravitational acceleration, g = 9.81 m/s
^{2}

Applying the formula of free fall time, in terms of velocity:

- t = v/g
- t = 26/9.81
- t = 2.65 s

Therefore, the time taken by a piece of paper to reach the ground is **2.65 s**.

Applying the formula of free fall distance, in terms of velocity:

- h = v
^{2}/2 g - h = (26)
^{2}/(2 × 9.81) - h = 676/19.62
- h = 34.45 m

Therefore, the distance traveled by a piece of paper is **34.45 m**.

### Problem #4

A box is released from a tower and takes 4 seconds to reach the ground. Calculate the distance traveled by the box during its free fall. Assume a gravitational acceleration of g = 9.81 m/s^{2}.

**Solution**

Given data:

- Time taken by a box to reach the ground, t = 4 s
- Distance traveled by a box, h = ?
- Gravitational acceleration, g = 9.81 m/s
^{2}

Applying the formula of free fall distance, in terms of time:

- h = ½ × g t
^{2} - h = ½ × (9.81) × (4)
^{2} - h = ½ × 9.81 × 16
- h = 78.48 m

Therefore, the distance travelled by a box is **78.48 m**.

## Related

**Free fall equation**- Air resistance formula
- Terminal velocity equation
- Gravity equation
- Inertia formula
- Acceleration formula
- Momentum equation
- Mass formula
- Velocity formula
- Pressure equation
- Kinematic equations
- Friction equation

## External links

- Kinematic Equations and Free Fall – The Physics Classroom
- Free Fall Calculator – Omni Calculator
- 3.7: Free Fall – Physics LibreTexts
- Free Fall in Physics | Definition, Equation & Examples – Study.com
- Free Fall – University Physics Volume 1 – BCcampus Pressbooks
- Free Falling Object – NASA (.gov)
- Kinematic Equations for Objects in Free Fall – Saylor Academy
- Free Fall Motion: Explanation, Review, and Examples – Albert
- 3.5 Free Fall | University Physics Volume 1 – Lumen Learning
- Free Fall Formula – SoftSchools.com
- What are free fall equations? – Quora
- Derivation of the Freefall Equation – xaktly.com
- Free Falling Object: Equation & Diagram – Vaia
- Free Fall – PhysicsTutorials.org
- Freely falling objects – University of Tennessee, Knoxville
- Free Fall Calculator – CalcTool
- Free Fall (Physics): Definition, Formula, Problems & Solutions (w/ Examples) – Sciencing

Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.