A boy standing at the top of a tower of 20m of height drops a stone. Assuming `g=10ms^(-2)`, the velocity with which it hits the ground is :-

Options

A. 10.0 m/s
B. 20.0m/s
C. 40.0m/s
D. 5.0m/s

Explanation

To solve the problem of finding the velocity with which the stone hits the ground after being dropped from a height of 20 meters, we can use the kinematic equation for motion under constant acceleration. Here’s a step-by-step solution:

### Step 1: Identify the known values
- Height of the tower (displacement, \( s \)) = 20 m (downward)
- Initial velocity (\( u \)) = 0 m/s (since the stone is dropped)
- Acceleration due to gravity (\( g \)) = 10 m/s² (downward)

### Step 2: Choose the direction for displacement and acceleration
Since the stone is falling downward, we can take downward as the positive direction. Thus:
- Displacement \( s = +20 \) m
- Acceleration \( a = +10 \) m/s²

### Step 3: Use the kinematic equation
We can use the following kinematic equation to find the final velocity (\( v \)):
\[ v^2 = u^2 + 2as \]
Substituting the known values into the equation:
- \( u = 0 \) m/s
- \( a = 10 \) m/s²
- \( s = 20 \) m

The equation becomes:
\[ v^2 = 0^2 + 2 \times 10 \times 20 \]
\[ v^2 = 0 + 400 \]
\[ v^2 = 400 \]

### Step 4: Calculate the final velocity
Now, take the square root of both sides to find \( v \):
\[ v = \sqrt{400} = 20 \text{ m/s} \]

### Step 5: Conclusion
The velocity with which the stone hits the ground is **20 m/s**.