Two stones are thrown up simultaneously from the edge of a cliff ` 240 m ` high with initial speed of ` 10 m//s and 40 m//s` respectively . Which of the following graph best represents the time variation of relative position of the speed stone with respect to the first ?
( Assume stones do not rebound after hitting the groumd and neglect air resistance , take ` g = 10 m//s^(2))`
( The figure are schematic and not drawn to scale )

Options

A.
B.
C.
D.

Explanation

To solve the problem of two stones thrown from a cliff, we need to analyze their motion step by step.

### Step 1: Understand the problem
We have two stones thrown from a height of 240 m. Stone A is thrown with an initial speed of 10 m/s, and Stone B is thrown with an initial speed of 40 m/s. We need to find the time variation of the relative position of Stone B with respect to Stone A.

### Step 2: Calculate the time taken for each stone to reach the highest point
For Stone A (initial speed = 10 m/s):
- The time taken to reach the highest point (where final velocity = 0):
\[ v = u - gt \implies 0 = 10 - 10t \implies t = 1 \text{ second} \]

For Stone B (initial speed = 40 m/s):
- The time taken to reach the highest point:
\[ v = u - gt \implies 0 = 40 - 10t \implies t = 4 \text{ seconds} \]

### Step 3: Calculate the maximum height reached by each stone
Using the formula for displacement:
\[ s = ut + \frac{1}{2}at^2 \]

For Stone A:
\[ s_A = 10 \times 1 + \frac{1}{2}(-10)(1^2) = 10 - 5 = 5 \text{ m} \]

For Stone B:
\[ s_B = 40 \times 4 + \frac{1}{2}(-10)(4^2) = 160 - 80 = 80 \text{ m} \]

### Step 4: Calculate the total height of each stone from the ground
The total height from the ground:
- Height of Stone A at its maximum:
\[ H_A = 240 + 5 = 245 \text{ m} \]
- Height of Stone B at its maximum:
\[ H_B = 240 + 80 = 320 \text{ m} \]

### Step 5: Calculate the time taken for each stone to hit the ground
For Stone A (total time of flight):
- It goes up for 1 second and then comes down. The time to fall from 245 m:
\[ s = ut + \frac{1}{2}gt^2 \implies 0 = 245 - \frac{1}{2}(10)t^2 \implies t^2 = 49 \implies t = 7 \text{ seconds} \]
Total time for Stone A = 1 second (up) + 7 seconds (down) = 8 seconds.

For Stone B (total time of flight):
- It goes up for 4 seconds and then comes down. The time to fall from 320 m:
\[ s = ut + \frac{1}{2}gt^2 \implies 0 = 320 - \frac{1}{2}(10)t^2 \implies t^2 = 64 \implies t = 8 \text{ seconds} \]
Total time for Stone B = 4 seconds (up) + 8 seconds (down) = 12 seconds.

### Step 6: Determine the relative position of Stone B with respect to Stone A
The relative position \(Y_B - Y_A\) can be calculated during their flight.

For \(t < 1\) (both stones are still rising):
\[ Y_A = 10t - 5t^2 \]
\[ Y_B = 40t - 5t^2 \]
\[ Y_B - Y_A = (40t - 5t^2) - (10t - 5t^2) = 30t \]

For \(1 < t < 4\) (Stone A is falling, Stone B is still rising):
\[ Y_A = 245 - 5(t - 1)^2 \]
\[ Y_B = 40t - 5t^2 \]
The relative position will change as Stone A falls.

For \(t > 4\) (Stone A is falling, Stone B is falling):
The relative position will be calculated based on their respective downward motions.

### Step 7: Analyze the graphs
The graph representing the relative position will show a linear increase during the first second (when both are rising), a change in slope when Stone A starts falling, and then a parabolic shape as Stone B continues to fall after Stone A has hit the ground.

### Conclusion
The graph that best represents the time variation of the relative position of Stone B with respect to Stone A will show a linear increase followed by a change in slope and then a parabolic curve after Stone A hits the ground.