A ball rleased from the tope of a tower travels ` (11) /( 36) ` of the height of the tower in the last second of its journey. The height of the tower is (g= 10 ms^(2)).

Options

A. ` 11 m`
B. (b) ` 36 m`
C. (c ) ` 18 0 m`
D. (d) ` ` 110 m`

Explanation

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

### Step 1: Understanding the problem
We are given that a ball is released from the top of a tower and travels \( \frac{11}{36} \) of the height of the tower in the last second of its journey. We need to find the height of the tower \( H \).

### Step 2: Setting up the equations
1. **Distance covered in the last second**: The distance covered by the ball in the last second can be expressed as:
\[ S_T = U + \frac{A}{2} (2T - 1) \]
where:
- \( S_T \) is the distance covered in the last second,
- \( U \) is the initial velocity (which is 0 since the ball is dropped),
- \( A \) is the acceleration due to gravity (given as \( g = 10 \, \text{m/s}^2 \)),
- \( T \) is the total time of fall.

Substituting the values, we have:
\[ S_T = 0 + \frac{10}{2} (2T - 1) = 5(2T - 1) \]

2. **Relating it to the height of the tower**: The distance covered in the last second is also given as:
\[ S_T = \frac{11}{36} H \]
Therefore, we can set the two expressions for \( S_T \) equal to each other:
\[ \frac{11}{36} H = 5(2T - 1) \]

### Step 3: Finding the total height
1. **Total height equation**: The total height \( H \) of the tower can also be expressed using the formula for distance traveled under constant acceleration:
\[ H = U T + \frac{1}{2} A T^2 \]
Since \( U = 0 \):
\[ H = \frac{1}{2} g T^2 = \frac{1}{2} \times 10 \times T^2 = 5T^2 \]

### Step 4: Substituting and solving for \( T \)
Now we have two equations:
1. \( \frac{11}{36} H = 5(2T - 1) \)
2. \( H = 5T^2 \)

Substituting \( H \) from the second equation into the first:
\[ \frac{11}{36} (5T^2) = 5(2T - 1) \]
This simplifies to:
\[ \frac{55}{36} T^2 = 10T - 5 \]
Multiplying through by 36 to eliminate the fraction:
\[ 55T^2 = 360T - 180 \]
Rearranging gives us:
\[ 55T^2 - 360T + 180 = 0 \]

### Step 5: Solving the quadratic equation
Using the quadratic formula \( T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 55 \), \( b = -360 \), and \( c = 180 \).
\[ T = \frac{360 \pm \sqrt{(-360)^2 - 4 \cdot 55 \cdot 180}}{2 \cdot 55} \]
Calculating the discriminant:
\[ (-360)^2 = 129600, \quad 4 \cdot 55 \cdot 180 = 39600 \]
Thus,
\[ T = \frac{360 \pm \sqrt{129600 - 39600}}{110} = \frac{360 \pm \sqrt{90000}}{110} = \frac{360 \pm 300}{110} \]
Calculating the two possible values for \( T \):
1. \( T = \frac{660}{110} = 6 \) seconds
2. \( T = \frac{60}{110} = \frac{6}{11} \) seconds

Since the time must be greater than 1 second (as it is the total time of fall), we take \( T = 6 \) seconds.

### Step 6: Finding the height \( H \)
Now substituting \( T = 6 \) seconds back into the height equation:
\[ H = 5T^2 = 5 \times 6^2 = 5 \times 36 = 180 \, \text{meters} \]

### Final Answer
The height of the tower is \( H = 180 \, \text{meters} \).