A stone dropped form the top of a tower is found to travel `(5//9)` of the height of tower during last second of its fall. The time of fall is :

Options

A. 2 s
B. 3 s
C. 4 s
D. 5 s

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the problem
A stone is dropped from the top of a tower. It travels \( \frac{5}{9} \) of the height of the tower during the last second of its fall. We need to find the total time of fall \( t \).

### Step 2: Define variables
Let:
- \( h \) = height of the tower
- \( t \) = total time of fall
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))

### Step 3: Distance traveled in the last second
The distance traveled during the last second of the fall can be expressed as:
\[ h' = \frac{5}{9} h \]
where \( h' \) is the distance traveled in the last second.

### Step 4: Use the equations of motion
1. The total distance fallen in \( t \) seconds is given by:
\[ h = \frac{1}{2} g t^2 \]
2. The distance fallen in the last second (from \( t-1 \) to \( t \)) can be calculated as:
\[ h' = \text{Distance fallen in } t \text{ seconds} - \text{Distance fallen in } (t-1) \text{ seconds} \]
\[ h' = \frac{1}{2} g t^2 - \frac{1}{2} g (t-1)^2 \]

### Step 5: Simplify the expression for \( h' \)
Expanding \( \frac{1}{2} g (t-1)^2 \):
\[ \frac{1}{2} g (t-1)^2 = \frac{1}{2} g (t^2 - 2t + 1) = \frac{1}{2} g t^2 - g t + \frac{1}{2} g \]
Thus,
\[ h' = \frac{1}{2} g t^2 - \left( \frac{1}{2} g t^2 - g t + \frac{1}{2} g \right) \]
\[ h' = g t - \frac{1}{2} g \]

### Step 6: Set up the equation
Now we can set up the equation:
\[ g t - \frac{1}{2} g = \frac{5}{9} h \]
Substituting \( h = \frac{1}{2} g t^2 \):
\[ g t - \frac{1}{2} g = \frac{5}{9} \left( \frac{1}{2} g t^2 \right) \]
\[ g t - \frac{1}{2} g = \frac{5}{18} g t^2 \]

### Step 7: Cancel \( g \) and rearrange
Dividing through by \( g \) (assuming \( g \neq 0 \)):
\[ t - \frac{1}{2} = \frac{5}{18} t^2 \]
Rearranging gives:
\[ \frac{5}{18} t^2 - t + \frac{1}{2} = 0 \]

### Step 8: Multiply through by 18 to eliminate the fraction
\[ 5t^2 - 18t + 9 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 5, b = -18, c = 9 \):
\[ t = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 5 \cdot 9}}{2 \cdot 5} \]
\[ t = \frac{18 \pm \sqrt{324 - 180}}{10} \]
\[ t = \frac{18 \pm \sqrt{144}}{10} \]
\[ t = \frac{18 \pm 12}{10} \]

Calculating the two possible values:
1. \( t = \frac{30}{10} = 3 \)
2. \( t = \frac{6}{10} = 0.6 \) (not physically meaningful in this context)

### Final Answer
Thus, the total time of fall is:
\[ \boxed{3 \text{ seconds}} \]