A body is dropped from a tower with zero velocity, reaches ground in 4s. The height of the tower is about:-

Options

A. 80m
B. 20m
C. 160m
D. 40m

Explanation

To solve the problem of finding the height of the tower from which a body is dropped, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the known values
- Initial velocity (u) = 0 m/s (since the body is dropped)
- Time (t) = 4 seconds
- Acceleration due to gravity (g) = 9.8 m/s² (approximately)

### Step 2: Use the equation of motion
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} g t^2 \]
Where:
- \( s \) = distance (height of the tower)
- \( u \) = initial velocity
- \( g \) = acceleration due to gravity
- \( t \) = time

### Step 3: Substitute the known values into the equation
Since \( u = 0 \):
\[ s = 0 \cdot t + \frac{1}{2} g t^2 \]
This simplifies to:
\[ s = \frac{1}{2} g t^2 \]

### Step 4: Calculate \( t^2 \)
Calculate \( t^2 \):
\[ t^2 = (4 \, \text{s})^2 = 16 \, \text{s}^2 \]

### Step 5: Substitute \( g \) and \( t^2 \) into the equation
Now substitute \( g = 9.8 \, \text{m/s}^2 \) and \( t^2 = 16 \, \text{s}^2 \):
\[ s = \frac{1}{2} \cdot 9.8 \cdot 16 \]

### Step 6: Calculate \( s \)
Calculate:
\[ s = \frac{1}{2} \cdot 9.8 \cdot 16 = 4.9 \cdot 16 \]
\[ s = 78.4 \, \text{m} \]

### Step 7: Round the answer
The height of the tower is approximately:
\[ s \approx 80 \, \text{m} \]

### Conclusion
Thus, the height of the tower is about 80 meters.

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