Calculating Distance Covered by a Freely Falling Body

A freely falling body near the surface of the Earth is subject to a constant acceleration due to gravity. This acceleration, denoted as g, is approximately 9.81 m/s2. In this article, we will use the equations of motion to determine the distance covered by a freely falling body in a given time interval.

The Equations of Motion

The primary equations of motion under constant acceleration are used to describe the motion of the freely falling body. These equations are essential for understanding the distance covered in any given time interval. The formula for the distance covered in the nth second is given by:

dn ? g (2n - 1)

Step-by-Step Calculation

Step 1: Calculate the Distance Covered in the 3rd Second

Given that the body travels 10 meters in the 3rd second:

d3 ? g (2 times; 3 - 1)

Substituting g 9.81 m/s2 and simplifying:

10 ? times; 9.81 times; (2 times; 3 - 1)

10 4.905 times; (6 - 1)

10 4.905 times; 5

This confirms our value for g is approximately 9.81 m/s2.

Step 2: Calculate the Distance Covered in the 4th Second

Using the same formula to find the distance covered in the 4th second:

d4 ? g (2 times; 4 - 1)

Substituting g 9.81 m/s2 and simplifying:

d4 ? times; 9.81 times; (2 times; 4 - 1)

d4 4.905 times; (8 - 1)

d4 4.905 times; 7

d4 34.335 ≈ 34.3 m

Conclusion

The distance covered by the freely falling body in the 4th second is approximately 34.3 meters.

A More Comprehensive Approach

Another approach to understanding the motion of a freely falling body uses the standard gravity formula:

h(t) ? g t2

Given that the initial velocity is 0 m/s, the distance covered in the first 3 seconds is:

h(3) ? times; 9.81 times; 32

h(3) 4.905 times; 9

h(3) 44.145 ≈ 44.1 m

This calculation provides the distance covered from the start to the 3rd second. Therefore, the distance covered in the 4th second can be found by subtracting the distance covered in the first 3 seconds from the total distance covered in 4 seconds:

h(4) ? times; 9.81 times; 42

h(4) 4.905 times; 16

h(4) 78.48 ≈ 78.5 m

Distance in the 4th second h(4) - h(3)

Distance in the 4th second 78.5 - 44.1 34.4 m

Summary

In summary, the distance covered by a freely falling body in the 4th second is approximately 34.4 meters. This approach uses the equations of motion under constant acceleration, which is fundamental to understanding the motion of freely falling bodies near the Earth's surface.

Key Points

The distance covered in the nth second is given by dn ? g (2n - 1). The standard gravity formula is h(t) ? g t2. Under free fall, the initial velocity is 0 m/s.

Conclusion

The distance covered in the next second by a freely falling body can be calculated using the equations of motion. Understanding these concepts is crucial for many areas of physics and engineering, particularly in problems involving motion under constant acceleration.