A stone falls freely from rest from aheight `h` and it travels a distance `9h//25` in the last second. The value of `h` is

Options

A. 145 m
B. 100 m
C. 125 m
D. 200 m

Explanation

To solve the problem step by step, we will follow the physics principles related to free fall and derive the required height \( h \).

### Step 1: Understand the problem
A stone falls freely from rest from a height \( h \). It travels a distance of \( \frac{9h}{25} \) in the last second of its fall. We need to find the value of \( h \).

### Step 2: Define the variables
Let \( n \) be the total time taken for the stone to fall from height \( h \) to the ground. The distance traveled in the last second (from \( n-1 \) seconds to \( n \) seconds) is given by:
\[ S_n = u + \frac{a}{2} (2n - 1) \]
Where:
- \( u = 0 \) (initial velocity, since it falls from rest)
- \( a = g \) (acceleration due to gravity)

Thus, the formula simplifies to:
\[ S_n = \frac{g}{2} (2n - 1) \]
Given that this distance is \( \frac{9h}{25} \), we can write:
\[ \frac{g}{2} (2n - 1) = \frac{9h}{25} \]

### Step 3: Use the formula for height
The total distance fallen from height \( h \) in \( n \) seconds is given by:
\[ h = ut + \frac{1}{2} g t^2 \]
Since \( u = 0 \):
\[ h = \frac{1}{2} g n^2 \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ h = 5n^2 \]

### Step 4: Substitute \( h \) in the distance equation
Now, we can substitute \( h \) from the height equation into the distance equation:
\[ \frac{10}{2} (2n - 1) = \frac{9 \cdot 5n^2}{25} \]
This simplifies to:
\[ 5(2n - 1) = \frac{45n^2}{25} \]
\[ 5(2n - 1) = \frac{9n^2}{5} \]

### Step 5: Clear the fraction
Multiply both sides by 5 to eliminate the fraction:
\[ 25(2n - 1) = 9n^2 \]
Expanding gives:
\[ 50n - 25 = 9n^2 \]
Rearranging the equation:
\[ 9n^2 - 50n + 25 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 9, b = -50, c = 25 \):
\[ n = \frac{50 \pm \sqrt{(-50)^2 - 4 \cdot 9 \cdot 25}}{2 \cdot 9} \]
Calculating the discriminant:
\[ n = \frac{50 \pm \sqrt{2500 - 900}}{18} \]
\[ n = \frac{50 \pm \sqrt{1600}}{18} \]
\[ n = \frac{50 \pm 40}{18} \]
Calculating the two possible values for \( n \):
1. \( n = \frac{90}{18} = 5 \)
2. \( n = \frac{10}{18} = \frac{5}{9} \) (not possible since time cannot be less than 1 second)

Thus, \( n = 5 \) seconds.

### Step 7: Find the height \( h \)
Now substitute \( n \) back into the height equation:
\[ h = 5n^2 = 5 \cdot 5^2 = 5 \cdot 25 = 125 \, \text{meters} \]

### Final Answer
The value of \( h \) is \( 125 \, \text{meters} \).

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