A boy standing at the top of a tower of `20 m`. Height drops a stone. Assuming `g = 10 ms^-2` towards north. The average acceleration of the body is.

Options

A. `20 m//s`
B. `40 m//s`
C. `5 m//s`
D. `10 m//s`

Answer

To solve the problem of finding the average acceleration of a stone dropped from a height of 20 meters, we can follow these steps:

### Step 1: Identify the known values
- Height of the tower (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 will use the second equation of motion, which relates the final velocity (v), initial velocity (u), acceleration (a), and distance (s):
\[ v^2 = u^2 + 2as \]

Here, since the stone is falling freely under gravity, we can substitute:
- \( a = g = 10 \, \text{m/s}^2 \)
- \( s = h = 20 \, \text{m} \)

### Step 3: Substitute the values into the equation
Substituting the known values into the equation:
\[ v^2 = 0^2 + 2 \times 10 \times 20 \]
\[ v^2 = 0 + 400 \]
\[ v^2 = 400 \]

### Step 4: Solve for the final velocity (v)
Taking the square root of both sides:
\[ v = \sqrt{400} \]
\[ v = 20 \, \text{m/s} \]

### Step 5: Determine the average acceleration
Since the stone is in free fall, the average acceleration is equal to the acceleration due to gravity:
\[ \text{Average acceleration} = g = 10 \, \text{m/s}^2 \]

### Final Answer
The average acceleration of the body is **10 m/s²**.

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