# Free fall equation

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 t2
• Velocity: v = g t

The distance traveled by the object during free fall can be calculated using the equation h = ½ × g t2. 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 = v2/2 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 = v2/2 g.

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## 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/s2.

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/s2

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

• h = ½ × g t2
• 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/s2.

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/s2

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 10th 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/s2.

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/s2

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 = v2/2 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/s2.

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/s2

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

• h = ½ × g t2
• 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.