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. `20 ms ^(-1)`
B. `40ms^(-1)`
C. `5 ms ^(-1)`
D. `10 ms ^(-1)`

Answer

To solve the problem of finding the velocity with which a stone hits the ground when dropped from a height of 20 meters, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the known values
- Height (h) = 20 m
- Initial velocity (u) = 0 m/s (since the stone is dropped)
- Acceleration due to gravity (g) = 10 m/s² (acting downwards)

### Step 2: Use the equation of motion
We can use the following equation of motion to find the final velocity (v) just before the stone hits the ground:

\[ v^2 = u^2 + 2gh \]

### Step 3: Substitute the known values
Since the initial velocity (u) is 0, the equation simplifies to:

\[ v^2 = 0 + 2gh \]

Substituting the values of g and h:

\[ v^2 = 2 \times 10 \, \text{m/s}^2 \times 20 \, \text{m} \]

### Step 4: Calculate the value
Now calculate:

\[ v^2 = 2 \times 10 \times 20 \]
\[ v^2 = 400 \]

### Step 5: Take the square root
To find v, take the square root of both sides:

\[ v = \sqrt{400} \]
\[ v = 20 \, \text{m/s} \]

### Final Answer
The velocity with which the stone hits the ground is **20 m/s**.

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