A stone is shot straight upward with a speed of 20 m/sec from a tower 200 m high. The speed with which it strikes the ground is approximately

Options

A. `60 m//sec`
B. `65m//sec`
C. `70m//sec`
D. `75m//sec`

Answer

To solve the problem of determining the speed with which the stone strikes the ground, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial velocity of the stone, \( u = 20 \, \text{m/s} \) (upward)
- Height of the tower, \( h = 200 \, \text{m} \) (downward)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (downward)

2. **Set Up the Equation:**
- We will use the third equation of motion, which is:
\[ v^2 = u^2 + 2as \]
- Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the displacement.

3. **Determine the Sign of Each Quantity:**
- Since the stone is shot upward, we take upward as positive. Therefore:
- \( u = +20 \, \text{m/s} \)
- The displacement \( s \) when the stone hits the ground is \( -200 \, \text{m} \) (since it moves downward).
- The acceleration \( a \) due to gravity is \( -g = -10 \, \text{m/s}^2 \).

4. **Substitute the Values into the Equation:**
- Substitute \( u \), \( a \), and \( s \) into the equation:
\[ v^2 = (20)^2 + 2(-10)(-200) \]
- Simplifying this gives:
\[ v^2 = 400 + 4000 \]
- Thus:
\[ v^2 = 4400 \]

5. **Calculate the Final Velocity:**
- Taking the square root to find \( v \):
\[ v = \sqrt{4400} = \sqrt{44 \times 100} = 10\sqrt{44} \]
- Further simplifying \( \sqrt{44} \):
\[ \sqrt{44} = \sqrt{4 \times 11} = 2\sqrt{11} \]
- Therefore:
\[ v = 20\sqrt{11} \, \text{m/s} \]

6. **Approximate the Value:**
- Using \( \sqrt{11} \approx 3.32 \):
\[ v \approx 20 \times 3.32 \approx 66.4 \, \text{m/s} \]

### Final Answer:
The speed with which the stone strikes the ground is approximately \( 66.4 \, \text{m/s} \).

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