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 of the two stones thrown from a cliff, we need to analyze their motions and find the relative position of the second stone with respect to the first stone over time.

### Step 1: Determine the equations of motion for both stones.

**For Stone 1 (initial speed = 10 m/s):**
- Initial height (h) = 240 m
- Initial velocity (u₁) = 10 m/s (upward)
- Acceleration (a) = -g = -10 m/s² (downward)

Using the equation of motion:
\[ s_1 = u_1 t + \frac{1}{2} a t^2 \]
Substituting the values:
\[ s_1 = 10t - \frac{1}{2} \cdot 10 \cdot t^2 \]
\[ s_1 = 10t - 5t^2 \]

**For Stone 2 (initial speed = 40 m/s):**
- Initial velocity (u₂) = 40 m/s (upward)

Using the same equation of motion:
\[ s_2 = u_2 t + \frac{1}{2} a t^2 \]
Substituting the values:
\[ s_2 = 40t - 5t^2 \]

### Step 2: Calculate the time taken for each stone to hit the ground.

**For Stone 1:**
Setting the displacement equal to the height of the cliff:
\[ 240 = 10t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 10t - 240 = 0 \]
Dividing by 5:
\[ t^2 - 2t - 48 = 0 \]
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -2, c = -48 \):
\[ t = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-48)}}{2 \cdot 1} \]
\[ t = \frac{2 \pm \sqrt{4 + 192}}{2} \]
\[ t = \frac{2 \pm 14}{2} \]
The positive root gives:
\[ t = 8 \text{ seconds} \]

**For Stone 2:**
Setting the displacement equal to the height of the cliff:
\[ 240 = 40t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 40t + 240 = 0 \]
Dividing by 5:
\[ t^2 - 8t + 48 = 0 \]
Using the quadratic formula:
\[ t = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 48}}{2 \cdot 1} \]
\[ t = \frac{8 \pm \sqrt{64 - 192}}{2} \]
This gives no real solution, indicating that Stone 2 will also hit the ground after 8 seconds.

### Step 3: Determine the relative position of Stone 2 with respect to Stone 1.

The relative position \( Y \) of Stone 2 with respect to Stone 1 is given by:
\[ Y = s_2 - s_1 \]
Substituting the equations:
\[ Y = (40t - 5t^2) - (10t - 5t^2) \]
\[ Y = 40t - 10t \]
\[ Y = 30t \]

### Step 4: Analyze the graph of relative position over time.

- From \( t = 0 \) to \( t = 8 \) seconds, the relative position \( Y \) increases linearly as \( 30t \).
- After \( t = 8 \) seconds, Stone 1 hits the ground, and Stone 2 continues to move upward until it reaches its peak and then falls back down.

### Conclusion:

The best representation of the time variation of the relative position of the second stone with respect to the first stone will show a linear increase until \( t = 8 \) seconds, followed by a downward parabolic motion after Stone 1 hits the ground.