Q 3.26 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.

As both the stones are being accelerated due to gravity its effect will come on the relative motion only when one of them reaches the ground. Till that point of time, the relative velocity of the second stone would remain the same with respect to the first stone.

Let us consider the upward direction to be positive.

V= 15 m s-1

V= 30 m s-1

Vrel = V2 -V= 30 - 15 = 15 m s-1

Initial velocity of the first stone(u) = 15 m s-1

Displacement from the point it has been thrown to the final point in its motion(s) = 200 m

Acceleration(a) = -g = -10 m s-2

Using the second equation of motion we have

$\\s=ut+\frac{1}{2}at^{2}\\ -200=15t-5t^{2}\\ t^{2}-3t-40=0\\ t^{2}-8t+5t-40=0\\ t(t-8)+5(t-8)=0\\ (t-8)(t+5)=0\\ t=8\ or\ t=-5$

As time cannot be negative the correct value of t is 8 seconds

For this much time the relative distance changes with the relative velocity.

Maximum relative distance is

$\\d=V_{rel}\times t\\ d=15\times 8\\ d=120\ m$

The graph is, therefore, correct till 8 seconds.

After that, the second stone will be moving towards the first stone with acceleration g towards it and their relative distance will keep on decreasing from 120 m till it becomes zero.

The velocity of the second stone 8 seconds after it has been thrown can be calculated as follows using the first equation of motion

$\\v=u+at\\ v=30-8\times 10\\ v=-50 m s^{-1}$

The time taken by it to travel 120 m in the downwards direction can be calculated as follows using the second equation of motion

$\\s=ut+\frac{1}{2}at^{2}\\ -120=-50t-5t^{2}\\ t^{2}+10t-24=0\\ t^{2}+12t-2t-24=0\\ t(t+12)-2(t+12)=0\\ (t-2)(t+12)=0\\ t=2s\ or\ t=-12s$

as t has to be positive the correct answer is 2 seconds.

The relative distance will become zero after a total time of 10 seconds which is the case as shown in the graph and therefore the graph shown in Figure correctly represents the time variation of the relative position of the second stone with respect to the first.

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