Calculating the Distance Covered by a Freely Falling Body: A Step-by-Step Guide

Introduction

The study of physics often involves understanding the motion of objects under different conditions. One such scenario is a freely falling body, which is a body in free fall under the influence of gravity alone. This article will explore how to calculate the distance covered by such a body in a specific time interval using the acceleration due to gravity. We will also delve into the formulas and calculations involved to ensure a clear understanding of the concept.

Distance Covered by a Freely Falling Body

The distance covered by a freely falling body under constant acceleration due to gravity can be calculated using the following formula:

(s ut frac{1}{2}gt^2)

Where:

(s) is the distance covered. (u) is the initial velocity of the body. (g) is the acceleration due to gravity. (t) is the time in seconds.

Example Calculation

Let's consider a scenario where a freely falling body has an initial velocity of (u 0) m/s, an acceleration due to gravity of (g 9.81 ,text{m/s}^2), and the duration of fall is (t 2) seconds.

Using the formula, we can calculate:

(s ut frac{1}{2}gt^2 0 times 2 frac{1}{2} times 9.81 times 2^2)

(s 0 frac{1}{2} times 9.81 times 4 19.62 ,text{meters})

Therefore, the distance covered by a freely falling body in 2 seconds is approximately (19.62) meters.

Additional Examples

Let's consider another scenario where an object is dropped from an initial height of 50 meters and falls under the influence of gravity for 2 seconds:

(s ut frac{1}{2}gt^2)

(u 0), (g 9.81 ,text{m/s}^2), (t 2) seconds

(s 0 times 2 frac{1}{2} times 9.81 times 2^2 19.62 ,text{meters})

In this case, the distance traveled is 19.62 meters, and the object is not yet on the ground. To determine the total time taken, we can use the formula for the height:

(h frac{1}{2}gt^2)

(50 frac{1}{2} times 9.81 times t^2)

(t sqrt{frac{50 times 2}{9.81}} 3.19) seconds

Differential Equation Method

The motion of a freely falling body can also be described using calculus. The acceleration due to gravity is constant, which means the second derivative of position with respect to time is constant:

(frac{d^2x}{dt^2} -g)

Integrating both sides, we get the velocity:

(v -gt v_0)

Assuming the initial velocity (v_0 0), the equation simplifies to:

(v -gt)

Integrating again:

(x -frac{1}{2}gt^2 x_0)

Where (x_0) is the initial height. If (x_0 50) meters and (t 2) seconds, the position of the object is:

(x -frac{1}{2} times 9.81 times 2^2 50)

(x -frac{1}{2} times 9.81 times 4 50 30.3867) meters

The distance traveled from the initial height to its current position is:

(50 - 30.3867 19.6133) meters

Therefore, the distance traveled is approximately (19.61) meters.

Conclusion

This article provided a comprehensive guide on how to calculate the distance covered by a freely falling body using the standard formula for motion under constant acceleration due to gravity. By understanding and applying these formulas, one can accurately predict and analyze the motion of freely falling objects in the real world.