A stone dropped from the top of a tower reaches the ground in 3 s. The height of the tower is

Options

A. 18.6 m
B. 39.2 m
C. 44.1 m
D. 98 m

Answer

To find the height of the tower from which a stone is dropped and reaches the ground in 3 seconds, we can use the kinematic equation for uniformly accelerated motion. The equation we will use is:

\[ S = ut + \frac{1}{2} a t^2 \]

Where:
- \( S \) is the displacement (height of the tower in this case),
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( t \) is the time taken.

### Step-by-step solution:

1. **Identify the initial conditions:**
- The stone is dropped, so the initial velocity \( u = 0 \) m/s.
- The time \( t = 3 \) seconds.
- The acceleration \( a = g = 9.8 \) m/s² (acceleration due to gravity).

2. **Substitute the values into the equation:**
Since the stone is falling downwards, we can take the downward direction as positive. Therefore, we can ignore the negative sign in the equation for this case. The equation simplifies to:
\[ S = 0 \cdot t + \frac{1}{2} g t^2 \]

3. **Calculate the displacement:**
Substitute \( g = 9.8 \) m/s² and \( t = 3 \) s into the equation:
\[ S = \frac{1}{2} \cdot 9.8 \cdot (3^2) \]
\[ S = \frac{1}{2} \cdot 9.8 \cdot 9 \]
\[ S = 4.9 \cdot 9 \]
\[ S = 44.1 \text{ meters} \]

4. **Conclusion:**
The height of the tower is **44.1 meters**.