A body falls freely from the top of a tower. It covers 36% of the total height in the lkast second before striking the ground level. The height of the tower is

Options

A. 50 m
B. 75 m
C. 100 m
D. 125 m

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the Problem
A body falls freely from the top of a tower, covering 36% of the total height in the last second before hitting the ground. We need to find the height of the tower.

### Step 2: Define Variables
Let:
- \( H \) = total height of the tower
- \( T \) = total time taken to fall from the top to the ground

### Step 3: Use the Equation of Motion
For a freely falling body from rest, the distance fallen in time \( T \) is given by:
\[ H = \frac{1}{2} g T^2 \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 4: Determine Distance Covered in the Last Second
The distance covered in the last second (from \( T-1 \) to \( T \)) is given as 36% of the total height:
\[ \text{Distance in last second} = H - \text{Distance covered in } (T-1) \text{ seconds} \]
This can be expressed as:
\[ \text{Distance in last second} = H - \frac{1}{2} g (T-1)^2 \]
Setting this equal to \( 0.36H \):
\[ H - \frac{1}{2} g (T-1)^2 = 0.36H \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ H - 0.36H = \frac{1}{2} g (T-1)^2 \]
\[ 0.64H = \frac{1}{2} g (T-1)^2 \]

### Step 6: Substitute \( H \) from Step 3
From Step 3, we know:
\[ H = \frac{1}{2} g T^2 \]
Substituting this into the equation from Step 5:
\[ 0.64 \left(\frac{1}{2} g T^2\right) = \frac{1}{2} g (T-1)^2 \]
Cancelling \( \frac{1}{2} g \) from both sides (assuming \( g \neq 0 \)):
\[ 0.64 T^2 = (T-1)^2 \]

### Step 7: Expand and Rearrange
Expanding the right side:
\[ 0.64 T^2 = T^2 - 2T + 1 \]
Rearranging gives:
\[ 0.64 T^2 - T^2 + 2T - 1 = 0 \]
\[ -0.36 T^2 + 2T - 1 = 0 \]
Multiplying through by -1:
\[ 0.36 T^2 - 2T + 1 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 0.36, b = -2, c = 1 \):
\[ T = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 0.36 \cdot 1}}{2 \cdot 0.36} \]
\[ T = \frac{2 \pm \sqrt{4 - 1.44}}{0.72} \]
\[ T = \frac{2 \pm \sqrt{2.56}}{0.72} \]
\[ T = \frac{2 \pm 1.6}{0.72} \]
Calculating the two possible values for \( T \):
1. \( T = \frac{3.6}{0.72} = 5 \) seconds (valid)
2. \( T = \frac{0.4}{0.72} \) (not valid as time cannot be negative)

### Step 9: Calculate the Height \( H \)
Now substituting \( T = 5 \) seconds back into the equation for height:
\[ H = \frac{1}{2} g T^2 = \frac{1}{2} \cdot 10 \cdot 5^2 = \frac{1}{2} \cdot 10 \cdot 25 = 125 \, \text{meters} \]

### Final Answer
The height of the tower is \( H = 125 \, \text{meters} \).

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