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.

t = √2 h/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

- https://www.physicsclassroom.com/class/1DKin/Lesson-5/How-Fast-and-How-Far
- https://www.omnicalculator.com/physics/free-fall
- https://study.com/learn/lesson/free-fall-physics-equation-examples.html
- https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/free-fall-without-air-resistance/
- https://www.studysmarter.us/explanations/physics/kinematics-physics/free-falling-object/