Determine the Free Fall Height: A Last Second Distance Puzzle

In this article, we will explore a classic problem involving the motion of a freely falling body, focusing on a distance covered in the last second of its descent. We will utilize the equations of motion under a uniform acceleration due to gravity to solve for the height from which the body was initially dropped. This article is valuable for students and professionals in physics and engineering who are interested in understanding and applying the principles of motion under gravity.

Understanding the Problem

The problem statement is as follows: A freely falling body covers a distance of h/2 in the last second of its motion. We are to determine the height h from which the body was initially dropped.

Equations of Motion

The motion of a freely falling body can be described using the following equations:

General motion distance equation: s ut (1/2)gt^2 Distance covered in free fall starting from rest: s (1/2)gt^2

Where:

s distance covered u initial velocity (0 for free fall) g acceleration due to gravity, approximately 9.81 m/s^2 t time in seconds

Solving the Problem

We will solve the problem step by-step using the equations of motion.

Step 1: Total Distance Consideration

Let T be the total time of fall. The total distance h equals:

h (1/2)gT^2

Step 2: Distance Covered in the Last Second

The distance covered in the last second (from T-1 to T) can be calculated as:

s_T (1/2)gT^2

s_{T-1} (1/2)g(T-1)^2 (1/2)gT^2 - gT (1/2)g

Thus, the distance covered in the last second is:

s_T - s_{T-1} (1/2)gT^2 - ((1/2)gT^2 - gT (1/2)g)

s_T - s_{T-1} gT - (1/2)g

Step 3: Setting Up the Equation

According to the problem, this distance equals h/2:

gT - (1/2)g (1/2)h

Step 4: Substituting for h

From the total distance equation, we have:

h (1/2)gT^2

Substituting this into the first equation gives:

gT - (1/2)g (1/4)gT^2

Step 5: Simplification

Dividing through by g (assuming g ≠ 0):

T - (1/2) (1/4)T^2

Rearranging it yields:

(1/4)T^2 - T - (1/2) 0

Step 6: Solving the Quadratic Equation

Utilizing the quadratic formula:

T [-b ± sqrt{b^2 - 4ac}] / (2a)

With a 1/4, b -1, and c -1/2:

T [1 ± sqrt{1 - 4(1/4)(-1/2)}] / (2(1/4))

T [1 ± sqrt{1 1/2}] / (1/2)

T [1 ± sqrt{3/2}] / (1/2)

T 2 ± sqrt{6}

Taking the positive root, since time cannot be negative:

T 2 sqrt{6}

Step 7: Calculating h

Substituting T back into the height equation:

h (1/2)gT^2

h (1/2)(9.81)(2 sqrt{6})^2

h 4.905(4 4sqrt{6} 6)

h 4.905(10 4sqrt{6})

h 49.05 23.62 72.67 meters (approximately)

Conclusion

The height h from which the body falls is approximately 72.67 meters.