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 of a body falling freely from the top of a tower and covering 36% of the total height in the last second before striking the ground, we can follow these steps:

### Step-by-Step Solution:

1. **Define Variables**:
Let the total height of the tower be \( H \). According to the problem, the body covers 36% of the total height in the last second. Therefore, the distance covered in the last second is:
\[ d = 0.36H \]

2. **Calculate Distance Covered Before the Last Second**:
The distance covered before the last second (i.e., in \( T - 1 \) seconds) is:
\[ d' = H - d = H - 0.36H = 0.64H \]

3. **Use the Equation of Motion**:
The distance covered in \( T - 1 \) seconds can be expressed using the second equation of motion:
\[ d' = \frac{1}{2} g (T - 1)^2 \]
Substituting \( d' \):
\[ 0.64H = \frac{1}{2} g (T - 1)^2 \]

4. **Express Total Height Using Time \( T \)**:
The total height \( H \) can also be expressed using the second equation of motion for \( T \) seconds:
\[ H = \frac{1}{2} g T^2 \]

5. **Substitute \( H \) in the Distance Equation**:
Substitute \( H \) from the previous equation into the distance equation:
\[ 0.64 \left(\frac{1}{2} g T^2\right) = \frac{1}{2} g (T - 1)^2 \]
Simplifying this gives:
\[ 0.32 g T^2 = \frac{1}{2} g (T^2 - 2T + 1) \]

6. **Cancel \( g \) and Rearrange**:
Since \( g \) is common on both sides, we can cancel it:
\[ 0.32 T^2 = \frac{1}{2} (T^2 - 2T + 1) \]
Multiplying through by 2 to eliminate the fraction:
\[ 0.64 T^2 = T^2 - 2T + 1 \]
Rearranging gives:
\[ 0 = T^2 - 2T + 1 - 0.64 T^2 \]
\[ 0 = 0.36 T^2 - 2T + 1 \]

7. **Multiply by 100 to Simplify**:
To eliminate decimals, multiply the entire equation by 100:
\[ 36 T^2 - 200 T + 100 = 0 \]

8. **Use the Quadratic Formula**:
Applying the quadratic formula \( T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 36 \), \( b = -200 \), and \( c = 100 \):
\[ T = \frac{200 \pm \sqrt{(-200)^2 - 4 \cdot 36 \cdot 100}}{2 \cdot 36} \]
\[ T = \frac{200 \pm \sqrt{40000 - 14400}}{72} \]
\[ T = \frac{200 \pm \sqrt{25600}}{72} \]
\[ T = \frac{200 \pm 160}{72} \]
This gives two possible values for \( T \):
\[ T = \frac{360}{72} = 5 \quad \text{(valid time)} \]
\[ T = \frac{40}{72} \quad \text{(not valid since time cannot be negative)} \]

9. **Calculate the Height of the Tower**:
Now that we have \( T = 5 \) seconds, we can find the height \( H \):
\[ H = \frac{1}{2} g T^2 \]
Assuming \( g = 10 \, \text{m/s}^2 \):
\[ H = \frac{1}{2} \cdot 10 \cdot (5^2) = \frac{1}{2} \cdot 10 \cdot 25 = 125 \, \text{meters} \]

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

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