Question -
Answer -
For the first stone:
Given,
Acceleration, a =–g = – 10 m/s2
Initial velocity, uI = 15 m/s
Now, we know
s1 = s0 + u1t +(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
∴– 5t2 +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 thown at time t = 0, the negative sign is notpossible
∴t = 8 s
For second stone:
Given,
Acceleration, a =– g = – 10 m/s2
Initial velocity, uII = 30 m/s
We know,
s2 = s0 + uIIt + (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 (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 tothis linear relation between (s2 – 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 isgiven by the quadratic equation:
s2 – s1 = 200 + 30t – 5t2
Therefore, the equation of linear and curved path is given by :
s2 – s1 = 15t (Linear path)
s2 – s1 =200 + 30t – 5t2 (Curved path)