Two stones are through 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 graphs best represents the time variation of relative position of the second stone with respect to the first? Assume stones do not rebound after hitting the ground and neglect air resistance, take . ` g= 10 m//s^(2) ` (The figures are schematic and not drawn to scale)

Options

A.
B.
C.
D.

Explanation

To solve the problem step by step, we need to analyze the motion of both stones and their relative positions over time.

### Step 1: Understand the Initial Conditions
- **Height of the cliff**: 240 m
- **Initial speed of Stone 1 (S1)**: 10 m/s (upwards)
- **Initial speed of Stone 2 (S2)**: 40 m/s (upwards)
- **Acceleration due to gravity (g)**: 10 m/s² (downwards)

### Step 2: Determine the Time of Flight for Each Stone
To find out how long each stone takes to hit the ground, we can use the kinematic equation:
\[ h = v_i t + \frac{1}{2} a t^2 \]
Where:
- \( h \) = height (240 m)
- \( v_i \) = initial velocity
- \( a \) = acceleration (which is -g because it acts downwards)

For **Stone 1 (S1)**:
\[ 240 = 10t - \frac{1}{2}(10)t^2 \]
\[ 240 = 10t - 5t^2 \]
\[ 5t^2 - 10t + 240 = 0 \]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 5, b = -10, c = 240 \):
\[ t = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 5 \cdot 240}}{2 \cdot 5} \]
\[ t = \frac{10 \pm \sqrt{100 - 4800}}{10} \]
Since the discriminant is negative, Stone 1 will hit the ground before reaching the maximum height.

For **Stone 2 (S2)**:
\[ 240 = 40t - \frac{1}{2}(10)t^2 \]
\[ 240 = 40t - 5t^2 \]
\[ 5t^2 - 40t + 240 = 0 \]
Using the quadratic formula again:
Here, \( a = 5, b = -40, c = 240 \):
\[ t = \frac{40 \pm \sqrt{(-40)^2 - 4 \cdot 5 \cdot 240}}{2 \cdot 5} \]
\[ t = \frac{40 \pm \sqrt{1600 - 4800}}{10} \]
Again, the discriminant is negative, indicating that Stone 2 will also hit the ground.

### Step 3: Analyze Relative Motion
The relative velocity between the two stones is:
\[ v_{rel} = v_{S2} - v_{S1} = 40 - 10 = 30 \text{ m/s} \]
This means that Stone 2 is moving away from Stone 1 at a constant rate of 30 m/s until one of them hits the ground.

### Step 4: Determine the Position Function
The position of each stone can be described by:
- For Stone 1:
\[ y_1(t) = 240 + 10t - 5t^2 \]
- For Stone 2:
\[ y_2(t) = 240 + 40t - 5t^2 \]

### Step 5: Find the Relative Position
The relative position of Stone 2 with respect to Stone 1 is given by:
\[ y_{rel}(t) = y_2(t) - y_1(t) \]
\[ y_{rel}(t) = (240 + 40t - 5t^2) - (240 + 10t - 5t^2) \]
\[ y_{rel}(t) = 30t \]

### Step 6: Determine the Graph Behavior
- For the time before both stones hit the ground, the relative position increases linearly with a slope of 30.
- After Stone 1 hits the ground, it stops moving, and Stone 2 continues to fall under gravity, which means the relative position will start to decrease as Stone 2 falls.

### Conclusion
The graph of the relative position of Stone 2 with respect to Stone 1 will show a linear increase until Stone 1 hits the ground, followed by a curve that decreases as Stone 2 falls.