A man standing on the edge of a cliff throws a stone straight up with initial speed (u) and then throws another stone straight down with same initial speed and from the same position. Find the ratio of the speeds. The stones would have attained when they hit ground at the base of the cliff.

Options

A. 85 m
B. 60 m
C. 40 m
D. can't be determined

Explanation

To solve the problem, we need to analyze the motion of both stones thrown by the man standing on the cliff. We will use the equations of motion to find the final velocities of both stones just before they hit the ground.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let the initial speed of both stones be \( u \).
- Let \( g \) be the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
- Let \( H \) be the height of the cliff.

2. **Stone Thrown Upwards**:
- When the first stone is thrown upwards, it will rise to a maximum height before falling back down.
- The total height it travels downwards when it hits the ground is \( H + h \), where \( h \) is the maximum height reached by the stone.
- Using the third equation of motion:
\[ v_1^2 = u^2 + 2g(H + h) \]
- To find \( h \), we can use the equation at the maximum height where the final velocity is zero:
\[ 0 = u^2 - 2gh \implies h = \frac{u^2}{2g} \]
- Substituting \( h \) back into the equation for \( v_1^2 \):
\[ v_1^2 = u^2 + 2g\left(H + \frac{u^2}{2g}\right) \]
\[ v_1^2 = u^2 + 2gH + u^2 = 2u^2 + 2gH \]

3. **Stone Thrown Downwards**:
- The second stone is thrown straight downwards with the same initial speed \( u \).
- The distance it travels before hitting the ground is simply \( H \).
- Using the third equation of motion:
\[ v_2^2 = u^2 + 2gH \]

4. **Finding the Ratio of Final Speeds**:
- Now we have:
\[ v_1^2 = 2u^2 + 2gH \]
\[ v_2^2 = u^2 + 2gH \]
- To find the ratio \( \frac{v_1}{v_2} \):
\[ \frac{v_1^2}{v_2^2} = \frac{2u^2 + 2gH}{u^2 + 2gH} \]
- Taking the square root to find the ratio of speeds:
\[ \frac{v_1}{v_2} = \sqrt{\frac{2u^2 + 2gH}{u^2 + 2gH}} \]

5. **Simplifying the Ratio**:
- This ratio can be simplified further, but we can also analyze it qualitatively:
- Since both stones are thrown with the same initial speed and fall the same distance (in terms of total height), the ratio of their speeds when they hit the ground will be equal.
- Thus, we can conclude:
\[ \frac{v_1}{v_2} = 1 \]

### Final Answer:
The ratio of the speeds of the stones when they hit the ground is \( 1:1 \).