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Step-by-Step Solution
Step 1: Identify the given information
• Two stones (stone 1 and stone 2) are thrown upwards simultaneously from the top of a 240 m high cliff.
• Initial velocity of stone 1, $u_1 = 10\,\text{m/s}$ (upwards).
• Initial velocity of stone 2, $u_2 = 40\,\text{m/s}$ (upwards).
• Acceleration due to gravity, $g = 10\,\text{m/s}^2$ (downwards).
Step 2: Write the displacement equations for each stone
We measure upward displacement as positive. Let $y_1$ and $y_2$ be the displacements (from the initial point on top of the cliff) of stone 1 and stone 2, respectively, at time $t$. The standard kinematics equation for vertical motion is:
$y = ut - \tfrac{1}{2}gt^2,$
where $u$ is the initial velocity (upwards taken as positive) and $g = 10\,\text{m/s}^2$.
For stone 1:
$y_1 = 10t - 5t^2.$
For stone 2:
$y_2 = 40t - 5t^2.$
Step 3: Calculate the relative position of stone 2 with respect to stone 1
The relative displacement of stone 2 from stone 1 is given by $y_2 - y_1$:
$y_2 - y_1 = (40t - 5t^2) - (10t - 5t^2) = 30t.$
This linear relationship holds as long as both stones are in the air. We need to find out until what time both stones remain in the air together.
Step 4: Find the time when stone 1 hits the ground
Stone 1 reaches the ground when its displacement below the top is $-240\,\text{m}$ (taking upward as positive, downward displacement is negative). So,
$y_1 = 10t - 5t^2 = -240.$
Rearranging:
$-5t^2 + 10t + 240 = 0.$
Solving this quadratic gives $t = 8\,\text{s}$ (only the positive root is physically meaningful). Hence, stone 1 hits the ground at $t = 8\,\text{s}$.
Therefore, from $t = 0$ to $t = 8\,\text{s}$,
$y_2 - y_1 = 30t,$
which is a straight line with a constant slope of 30.
Step 5: Relative position for times after stone 1 hits the ground
After $t = 8\,\text{s}$, stone 1 remains on the ground at $y_1 = -240\,\text{m}$. Meanwhile, stone 2 continues moving until it also hits the ground. Let us keep the same reference (top of the cliff) for $y_2$:
$y_2 = 40t - 5t^2.$
Hence, for $t > 8\,\text{s}$,
$y_2 - y_1 = \bigl(40t - 5t^2\bigr) - \bigl(-240\bigr) = 40t - 5t^2 + 240.$
This expression is a quadratic in $t$ (parabola) for $t > 8\,\text{s}$.
Step 6: Identify the form of the resulting graph
From $t = 0$ to $t = 8\,\text{s}$, the relative position $y_2 - y_1$ is a straight line of slope 30 (initially starting from 0 at $t=0$).
From $t = 8\,\text{s}$ onwards, the relative position $y_2 - y_1 = 40t - 5t^2 + 240$ forms a parabola until stone 2 hits the ground. Because the coefficient of $t^2$ is $-5$ here, that portion of the displacement graph opens downward, but in the schematic representation (where we only plot the relative displacement of stone 2 with respect to stone 1 against $t$), the shape after $t = 8\,\text{s}$ will show a curvature indicative of a parabolic trend.
Step 7: Conclude and match the correct graphical option
Based on the above reasoning:
The plot is a straight line for $0 \leq t \leq 8\,\text{s}$, then changes to a parabola for $t > 8\,\text{s}$.
The correct graphical representation from the given options is:
