Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of $$15ms^{-1}$$ and $$30ms^{-1}$$Verify that the graph shown in given 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 = $$10ms^{-2}$$ Give the equations for the linear and curved parts of the plot.

Asked by Abhisek | 1 year ago |  290

##### Solution :-

For the first stone:

Given,

Acceleration, a = –g = – 10 m/s2
Initial velocity, uI = 15 m/s

Now, we know
s1 = s0 + u1t + ($$\dfrac{1}{2}$$)at2
Given, height of the tree, s0 = 200 m
s1 = 200 + 15t – 5t2      . . . . . . . . .  . ( 1 )
When this stone hits the jungle floor, s1 = 0
– 5t+ 15t + 200 = 0
t2 – 3t – 40 = 0
t2 – 8t + 5t – 40 = 0
t (t – 8) + 5 (t – 8) = 0
t = 8 s or t = – 5 s
Since, the stone was thrown at time t = 0, the negative sign is not possible
t = 8 s
For second stone:

Given,

Acceleration, a = – g = – 10 m/s2
Initial velocity, uII = 30 m/s

We know,
s2 = s0 + uIIt + ($$\dfrac{1}{2}$$)at2
= 200 + 30t – 5t2 . . . . . . . . .  . . .  . ( 2 )
when this stone hits the jungle floor; s2 = 0
– 5t2 + 30 t + 200 = 0
t2 – 6t – 40 = 0
t2 – 10t + 4t + 40 = 0
t (t – 10) + 4 (t – 10) = 0
(t – 10) (t + 4) = 0
t = 10 s or t = – 4 s
Here again, the negative sign is not possible
t = 10 s
Subtracting equations ( 1 ) from equation ( 2 ), we get
s2 – s1 = (200 + 30t -5t2) – (200 + 15t -5t2)
s2 – s1 =15t                                           . . .  . . . . . . . . . .. . . . . . ( 3 )
Equation ( 3 ) represents the linear trajectory of the two stone, because to this

linear relation between (s– s1) and t,,  the projection is a straight line till 8 s.
The maximum distance between the two stones is at t = 8 s.
(s2 – s1)max = 15× 8 = 120 m
This value has been depicted correctly in the above graph.
After 8 s, only the second stone is in motion whose variation with time is given by the quadratic equation:
s2 – s= 200 + 30t – 5t2
Therefore, the equation of linear and curved path is given by :
s– s1 = 15t (Linear path)

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

Answered by Pragya Singh | 1 year ago

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