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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 Fig. 3.27 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.
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 Fig. 3.27 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.
CBSE | Class 11 | Motion in a Straight Line | NCERT | Physics
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 Fig. 3.27 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.

Ans.

For the first stone, it is given that the acceleration is

a = –g = – 10 m/s2
And the initial velocity is uI = 15 m/s

Using the expression of equation of motion => s1 = s0 + u1t + (1/2)at2
where s0 represents the height of the tree = 200 m
On substituting values, we get => s1 = 200 + 15t – 5t2      . . . . . . . . .  . ( 1 )
When the stone hits the floor, s1 becomes zero.
Therefore, we get

– 5t2 + 15t + 200 = 0
Or, t2 – 3t – 40 = 0
Solving the above equation by middle term splitting method

we get => t2 – 8t + 5t – 40 = 0
t (t – 8) + 5 (t – 8) = 0
Thus, t = 8 s or t = – 5 s
And as the stone was thrown at time t = 0, the negative sign makes no sense so, t = 8 s
now, for the second stone it is given that  the acceleration is

a = – g = – 10 m/s2
And the nitial velocity, uII = 30 m/s

Similarly using the expression used in case 1, we can write
s2 = s0 + uIIt + (1/2)at2
Upon substituting values,

s2 = 200 + 30t – 5t2 . . . . . . . . .  . . .  . ( 2 )
when this stone hits jungle floor; s2 becomes 0
– 5t2 + 30 t + 200 = 0
t2 – 6t – 40 = 0
Factorizing by the middle term splitting method, we get => t2 – 10t + 4t + 40 = 0
t (t – 10) + 4 (t – 10) = 0
Or, (t – 10) (t + 4) = 0
Therefore, t = 10 s or t = – 4 s
Here again, the negative sign will not be considered
∴ t = 10 s
Subtracting (1) from (2), we get
s2 – s1 = (200 + 30t -5t2) – (200 + 15t -5t2)
s2 – s1 =15t                                           . . .  . . . . . . . . . .. . . . . . ( 3 )
Equation (3) shows the linear trajectory of the two stones considered here. This is because the projection is a straight line till 8 s for this linear relationship between (s2 – s1) and t.
The maximum distance between two stones occurs at t = 8 s.
So, for maximum distance we can write => (s2 – s1)max = 15× 8 = 120 m
This value has been shown correctly in the above graph.
Only the second stone is moving after 8 seconds, and its variation with time is described by the quadratic equation:
s2 – s1 = 200 + 30t – 5t2
Therefore, the equation of linear and curved becomes :
s2 – s1 = 15t (Linear path)

s2 ­– s1 = 200 + 30t – 5t2 (Curved path)

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