Two stones are thrown up simultaneously from the edge of a cliff ` 200 m` high with initial speeds of ` 15 ms?^(-1)` and ` 30^(-1)`. Verify that the graph shown in Fig. 2 ( NCT). 13 , correctly represents the time variation of the relativ e position of the second stone with respect to the first. Neglect the air resistance and assume that the stones do not rebound after hitting the ground. Taje ` g= 10 ms^(-2)`.Give equations for the linear and curved parts of the plot.
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Explanation

Taking vertical upward motion of the first stone for time (t), we have ltbRgt ` x_0 = 200 m, u= 15 m//s , a=- 10 m//s^@, t=t , x= x_1`
As, ` x=x_0 + ut + 1/2 at^2 :. X_1 = 200 + 15 t + 1/2 (-10) t^2 or x_1 = 2300 = 15 t- 5 t^2` …..(i0
Taking vertical upward motion of the second stone form time (t), we have
` x_0 = 200 m, u= 30 ms^(-1), a=- ms^(10) ms^-20 , t=t, x=x_2`
Then ` x_2 = 200 + 30 t - 1.2 xx 10 t^2 = 200 + 30 t- 5 t^2` ....(ii)`
:. Either ` t= 8 s or -5 s`
Since ` t=0` corresponds to the instanct, when the stone was projected . Hence negative time has no meaning in this case. So ` t= 8 s`. When the second stone hits the ground,x-2=0`, so
` 0= 200 + 30 t- 5 t^2` or t^2 - 6 t- 40 =0` or ` 9 t- 10 9t =4 ) =0`
Therfore, either `t= 10 s` or ` t=- 4 s`
Since ` t=- 4 s` meaningles, so ` t= 10 s`
Relative positionof second stone w.r.t. first is ` = x_2 -x_1 = 15 t` .....(ii)
[ From 9i0 and (ii) ]`
Since (x_2 -x_1) and (t0 are linearly related, therefore, the graph is a straight line till ` t= 8 s`.
For maximum separation, `t=8 s`, so maximu separation `= 15 xx 8 = 12 0 m`
After ` 8 second` only stone would be in motion for ` 2 secinds`, so the graph is in accordance with the quadratic equation,`x-2 = 200 + 30 t- 5 t^2` for the interval of time ` 8 seconds1 to `10 seconds`.