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 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 1: Understand the problem
Let the total height of the tower be \( h \). The body is dropped from the top, so the initial velocity \( u = 0 \). The distance covered in the last second of the fall is given as \( \frac{7}{16} h \).

### Step 2: Use the equations of motion
The distance covered in \( t \) seconds is given by the equation:
\[ h = ut + \frac{1}{2}gt^2 \]
Since \( u = 0 \), this simplifies to:
\[ h = \frac{1}{2}gt^2 \]

### Step 3: Distance covered in the last second
The distance covered in the last second (from \( t-1 \) to \( t \)) can be expressed as:
\[ \text{Distance in last second} = h - \text{Distance covered in } (t-1) \text{ seconds} \]
The distance covered in \( t-1 \) seconds is:
\[ h_{t-1} = \frac{1}{2}g(t-1)^2 \]
Thus, the distance covered in the last second is:
\[ \text{Distance in last second} = h - h_{t-1} = \frac{1}{2}gt^2 - \frac{1}{2}g(t-1)^2 \]

### Step 4: Set up the equation
Now we can set up the equation based on the information that the distance covered in the last second is \( \frac{7}{16}h \):
\[ \frac{1}{2}gt^2 - \frac{1}{2}g(t-1)^2 = \frac{7}{16}h \]
Substituting \( h = \frac{1}{2}gt^2 \) into the equation gives:
\[ \frac{1}{2}gt^2 - \frac{1}{2}g(t-1)^2 = \frac{7}{16} \left(\frac{1}{2}gt^2\right) \]

### Step 5: Simplify the equation
Expanding \( (t-1)^2 \):
\[ (t-1)^2 = t^2 - 2t + 1 \]
Thus,
\[ \frac{1}{2}gt^2 - \frac{1}{2}g(t^2 - 2t + 1) = \frac{7}{32}gt^2 \]
This simplifies to:
\[ \frac{1}{2}g(2t - 1) = \frac{7}{32}gt^2 \]
Cancelling \( g \) (assuming \( g \neq 0 \)):
\[ t - \frac{1}{2} = \frac{7}{16}t^2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ \frac{7}{16}t^2 - t + \frac{1}{2} = 0 \]
Multiplying through by 16 to eliminate the fraction:
\[ 7t^2 - 16t + 8 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here \( a = 7, b = -16, c = 8 \):
\[ t = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 7 \cdot 8}}{2 \cdot 7} \]
Calculating the discriminant:
\[ t = \frac{16 \pm \sqrt{256 - 224}}{14} = \frac{16 \pm \sqrt{32}}{14} = \frac{16 \pm 4\sqrt{2}}{14} \]
Simplifying gives:
\[ t = \frac{8 \pm 2\sqrt{2}}{7} \]

### Step 8: Finding the positive root
Since time cannot be negative, we take the positive root:
\[ t = \frac{8 + 2\sqrt{2}}{7} \]
Calculating this gives approximately \( t \approx 4 \) seconds.

### Conclusion
The time of fall \( t \) is approximately \( 4 \) seconds.