A stone is thrown vertically upward with a speed of `10.0 "ms"^(-1)` from the edge of a cliff 65 m high. What will be its speed just before hitting the bottom ?

Options

A. 3.14 m/s
B. 37.14 m/s
C. 13.71 m/s
D. 14.71 m/s

Answer

To solve the problem of finding the speed of the stone just before it hits the bottom after being thrown vertically upward from a height of 65 m, we can use the third equation of motion.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial velocity (u) = 10 m/s (upward)
- Height of the cliff (h) = 65 m (downward)
- Acceleration due to gravity (g) = 9.81 m/s² (downward)

2. **Choose the Direction:**
- We will take downward as positive for our calculations.

3. **Use the Third Equation of Motion:**
The third equation of motion is given by:
\[ v^2 = u^2 + 2gh \]
where:
- v = final velocity (just before hitting the ground)
- u = initial velocity
- g = acceleration due to gravity
- h = height

4. **Substitute the Values:**
Since the stone is thrown upward, we need to consider the upward motion and then the downward motion. The stone will first go up and then come down. However, we can directly calculate the final velocity just before hitting the ground using:
\[ v^2 = (10)^2 + 2 \cdot 9.81 \cdot 65 \]

5. **Calculate Each Term:**
- \( (10)^2 = 100 \)
- \( 2 \cdot 9.81 \cdot 65 = 1277.3 \)

6. **Combine the Results:**
\[ v^2 = 100 + 1277.3 = 1377.3 \]

7. **Take the Square Root:**
\[ v = \sqrt{1377.3} \approx 37.14 \, \text{m/s} \]

8. **Conclusion:**
The speed of the stone just before hitting the bottom is approximately **37.14 m/s**.