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Asked by Jitender Last Modified
Answer For first stone:
Initial velocity, uI = 15 m/s
Acceleration, a = –g = – 10 m/s2
Using the relation,
When this stone hits the ground, x1 = 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 projected at time t = 0, the negative sign before time is meaningless.
∴t = 8 s
For second stone:
Initial velocity, uII = 30 m/s
Acceleration, a = –g = – 10 m/s2
Using the relation,
At the moment when this stone hits the ground; x2 = 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 meaningless.
∴t = 10 s
Subtracting equations (i) and (ii), we get
Equation (iii) represents the linear path of both stones. Due to this linear relation between (x2 – x1) and t, the path remains a straight line till 8 s.
Maximum separation between the two stones is at t = 8 s.
(x2 – x1)max = 15× 8 = 120 m
This is in accordance with the given graph.
After 8 s, only second stone is in motion whose variation with time is given by the quadratic equation:
x2 – x1 = 200 + 30t – 5t2
Hence, the equation of linear and curved path is given by
x2 – x1 = 15t (Linear path)
x2 – x1 = 200 + 30t – 5t2 (Curved path)
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