Two stones are thrown up simultaneously from the edge of a cliff $200 \; m$ high with initial speeds of $15 \; m s^{-1}$ and $30 \; m s^{-1}$. Verify that the graph shown in Figure correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take $g = 10 \; m s^{-2}$. Give the equations for the linear and curved parts of the plot.

(N/A) For the first stone:
Initial velocity,$u_{1} = 15 \; m s^{-1}$
Acceleration,$a = -g = -10 \; m s^{-2}$
Using the equation of motion,$x_{1} = x_{0} + u_{1}t + \frac{1}{2}at^{2}$
Given the height of the cliff,$x_{0} = 200 \; m$,so $x_{1} = 200 + 15t - 5t^{2} \; \dots (i)$
When this stone hits the ground,$x_{1} = 0$,so $-5t^{2} + 15t + 200 = 0 \implies t^{2} - 3t - 40 = 0$.
Solving the quadratic equation,$(t - 8)(t + 5) = 0$. Since $t > 0$,$t = 8 \; s$.
For the second stone:
Initial velocity,$u_{2} = 30 \; m s^{-1}$
Acceleration,$a = -g = -10 \; m s^{-2}$
$x_{2} = 200 + 30t - 5t^{2} \; \dots (ii)$
When this stone hits the ground,$x_{2} = 0$,so $-5t^{2} + 30t + 200 = 0 \implies t^{2} - 6t - 40 = 0$.
Solving the quadratic equation,$(t - 10)(t + 4) = 0$. Since $t > 0$,$t = 10 \; s$.
For $0 \le t \le 8 \; s$,both stones are in the air:
$x_{2} - x_{1} = (200 + 30t - 5t^{2}) - (200 + 15t - 5t^{2}) = 15t$.
This is a linear equation representing the straight line part of the graph.
For $8 \; s < t \le 10 \; s$,only the second stone is in the air $(x_{1} = 0)$:
$x_{2} - x_{1} = x_{2} - 0 = 200 + 30t - 5t^{2}$.
This is a quadratic equation representing the curved part of the graph.