A body dropped from the top of a tower covers 7/16 of the total height in the last second of its fall. The time of fall is

Options

A. `2sec`
B. `4sec`
C. `1sec`
D. `((50)/(7))sec`

Explanation

To solve the problem, we need to find the total time of fall \( t \) for a body dropped from the top of a tower, given that it covers \( \frac{7}{16} \) of the total height in the last second of its fall.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let \( h \) be the total height of the tower.
- The distance covered in the last second of the fall is given as \( x = \frac{7}{16} h \).
- The time of fall is \( t \) seconds, and during the last second, the body falls from \( t-1 \) seconds to \( t \) seconds.

2. **Distance Covered in the Last Second**:
- The distance covered in the last second can be expressed as:
\[ x = h - \text{distance covered in } (t-1) \text{ seconds} \]
- The distance covered in \( (t-1) \) seconds can be calculated using the equation of motion:
\[ h - x = \frac{1}{2} g (t-1)^2 \]
- Substituting \( x = \frac{7}{16} h \):
\[ h - \frac{7}{16} h = \frac{1}{2} g (t-1)^2 \]
- This simplifies to:
\[ \frac{9}{16} h = \frac{1}{2} g (t-1)^2 \]

3. **Total Distance Fallen**:
- The total distance fallen \( h \) can also be expressed in terms of \( t \):
\[ h = \frac{1}{2} g t^2 \]

4. **Substituting for \( h \)**:
- From the second equation, we can express \( h \) as:
\[ h = \frac{g t^2}{2} \]
- Substitute this into the equation from step 2:
\[ \frac{9}{16} \left(\frac{g t^2}{2}\right) = \frac{1}{2} g (t-1)^2 \]
- Simplifying gives:
\[ \frac{9 g t^2}{32} = \frac{1}{2} g (t^2 - 2t + 1) \]

5. **Eliminating \( g \)**:
- Since \( g \) is common, we can cancel it out (assuming \( g \neq 0 \)):
\[ \frac{9 t^2}{32} = \frac{1}{2} (t^2 - 2t + 1) \]
- Multiplying through by 32 to eliminate the fraction:
\[ 9 t^2 = 16(t^2 - 2t + 1) \]

6. **Rearranging the Equation**:
- Expanding the right-hand side:
\[ 9 t^2 = 16t^2 - 32t + 16 \]
- Rearranging gives:
\[ 16t^2 - 9t^2 - 32t + 16 = 0 \]
\[ 7t^2 - 32t + 16 = 0 \]

7. **Using the Quadratic Formula**:
- We can solve this quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{32 \pm \sqrt{(-32)^2 - 4 \cdot 7 \cdot 16}}{2 \cdot 7} \]
- Calculate the discriminant:
\[ 1024 - 448 = 576 \]
- Thus:
\[ t = \frac{32 \pm 24}{14} \]
- This gives two possible values:
\[ t = \frac{56}{14} = 4 \quad \text{and} \quad t = \frac{8}{14} = \frac{4}{7} \]

8. **Final Answer**:
- The total time of fall is \( t = 4 \) seconds (the positive value).