For a free falling object, the **distance **(h) is equal to half times the product of **gravity** (g) and the square of **time **(t). Using the equation of free fall: **h = ½ × g t**** ^{2}**, the value of distance travelled by a free falling object can be calculated.

Here’s the equation of free fall:

- Free fall
**distance**and**velocity**in terms of time,**h = ½ × g t**and^{2}**v = g t**

- Free fall
**time**and**velocity**in terms of distance,**t = √2 h/g**and**t = √2 g h**

- Free fall
**time**and**distance**in terms of velocity,**t = v/g**and**h = v**^{2}**/2 g**

Let’s solve some problems based on this equation, so you’ll get a clear idea.

## Free Fall Practice Problems

**Problem 1:** One feather is allowed to free fall from the top of a building. Calculate the distance travelled by a feather, if the time taken by a feather to reach the ground is 20 s. Also calculate the velocity of a feather. (Take the value of gravitational acceleration, g = 9.81 m/s^{2})

Solution:

Given data:

Time taken by a feather to reach the ground, t = 20 s

Gravitational acceleration, g = 9.81 m/s^{2}

Distance travelled by a feather, h = ?

Velocity of a feather, v = ?

Using the equation of free fall, (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 travelled by a feather is **1962 m**.

Using the equation of free fall, (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:** One ball is allowed to free fall from the top of a tower. In order to reach the ground, a ball has to travel the distance of 600 m. After what time, a ball reaches the ground and what is the velocity of a ball? (Take the value of gravitational acceleration, g = 9.81 m/s^{2})

Solution:

Given data:

Distance travelled by a ball, h = 600 m

Gravitational acceleration, g = 9.81 m/s^{2}

Time taken by a ball to reach the ground, t = ?

Velocity of a ball, v = ?

Using the equation of free fall, (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**.

Using the equation of free fall, (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:** One small piece of paper is allowed to free fall from the 10^{th} floor of a building. If a piece of paper is falling with the velocity of 26 m/s, then how much time will it take to reach the ground? And what is the distance travelled by a piece of paper? (Take the value of gravitational acceleration, g = 9.81 m/s^{2})

Solution:

Given data:

Velocity of a piece of paper, v = 26 m/s

Time taken by a piece of paper to reach the ground, t = ?

Distance travelled by a piece of paper, h = ?

Gravitational acceleration, g = 9.81 m/s^{2}

Using the equation of free fall, (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**.

Using the equation of free fall, (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 travelled by a piece of paper is **34.45 m**.

**Problem 4:** When one box is allowed to free fall from a tower, it takes 4 s to reach the ground. Calculate the distance travelled by a box. (Take the value of gravitational acceleration, g = 9.81 m/s^{2})

Solution:

Given data:

Time taken by a box to reach the ground, t = 4 s

Distance travelled by a box, h = ?

Gravitational acceleration, g = 9.81 m/s^{2}

Using the equation of free fall, (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**.

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