A stone from the top of a tower, travels 35 m in the last second of its journey. The height of the tower is

Options

A. `20 m`
B. `40 m`
C. `60 m`
D. `80 m`

Explanation

To find the height of the tower from which a stone is dropped, given that it travels 35 m in the last second of its journey, we can follow these steps:

### Step 1: Understand the motion
The stone is dropped from the top of the tower, which means its initial velocity (u) is 0 m/s. The stone is under the influence of gravity, which provides a constant acceleration (a) of approximately 10 m/s² (we will use g = 10 m/s² for simplicity).

### Step 2: Use the formula for distance traveled in the nth second
The distance traveled in the nth second is given by the formula:
\[ d_n = u + \frac{a}{2} \cdot (2n - 1) \]
Here, \(d_n\) is the distance traveled in the nth second, \(u\) is the initial velocity, \(a\) is the acceleration, and \(n\) is the total time taken in seconds.

### Step 3: Substitute the known values
Since the stone is dropped, \(u = 0\) and \(a = g = 10 \, \text{m/s}^2\). We know that the stone travels 35 m in the last second, so we can set \(d_n = 35\):
\[ 35 = 0 + \frac{10}{2} \cdot (2n - 1) \]
This simplifies to:
\[ 35 = 5(2n - 1) \]

### Step 4: Solve for n
Now, we can solve for \(n\):
\[ 35 = 5(2n - 1) \]
Dividing both sides by 5:
\[ 7 = 2n - 1 \]
Adding 1 to both sides:
\[ 8 = 2n \]
Dividing by 2:
\[ n = 4 \]

### Step 5: Calculate the height of the tower
Now that we know the total time of flight \(n = 4\) seconds, we can calculate the height of the tower using the formula for distance traveled under constant acceleration:
\[ H = ut + \frac{1}{2} a t^2 \]
Substituting \(u = 0\), \(a = 10 \, \text{m/s}^2\), and \(t = 4\):
\[ H = 0 \cdot 4 + \frac{1}{2} \cdot 10 \cdot (4^2) \]
This simplifies to:
\[ H = \frac{1}{2} \cdot 10 \cdot 16 \]
\[ H = 5 \cdot 16 = 80 \, \text{m} \]

### Final Answer
The height of the tower is **80 meters**.

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