Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s. 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/s2. Give the equations for the linear and curved parts of the plot.
Answers
Explanation:
Answer:-
For first stone:
Initial velocity, uI = 15 m/s
Acceleration, a = –g = – 10 m/s2
Using the relation,
x1 = x0 + u1t + (1/2)at2
Where, height of the cliff, x0 = 200 m
x1 = 200 + 15t – 5t2 ……(i)
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,
x2 = x0 + uIIt + (1/2)at2
= 200 + 30t – 5t2……..(ii)
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
x2 – x1 = (200 + 30t -5t2) – (200 + 15t -5t2)
x2 – x1 =15t ……. (iii)
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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For first stone:
Initial velocity, u
1
=15m/s
Acceleration, a=−g=−10m/s
2
Using the relation,
x
1
=x
o
+u
1
t+
2
1
at
2
Where, height of the cliff, x
o
=200m
x
1
=200+15t−5t
2
.....(i)
When this stone hits the ground, x
1
=0
∴−5t
2
+15t+200=0
On solving we can get:
t=8s or t=–5s
Since the stone was projected at time t=0, the negative sign before time is meaningless.
∴t=8s
For second stone:
Initial velocity, u
2
=30m/s
Acceleration, a=−g=−10m/s
2
Using the relation,
x
2
=x
o
+u
2
t+
2
1
at
2
=200+30t−5t
2
........(ii)
At the moment when this stone hits the ground; x
2
=0
−5t
2
+30t+200=0
t=10s or t=−4s
Here again, the negative sign is meaningless.
∴t=10s
Subtracting equations (i) and (ii), we get
x
2
−x
1
=(200+30t−5t
2
)−(200+15t−5t
2
)
x
2
−x
1
=15t........(iii)
Equation (iii) represents the linear path of both stones. Due to this linear relation between (x
2
−x
1
) and t, the path remains a straight line till 8s.
Maximum separation between the two stones is at t=8s
(x
2
−x
1
)
max
=15×8=120m
This is in accordance with the given graph.
After 8s, only second stone is in motion whose variation with time is given by the quadratic equation:
x
2
−x
1
=200+30t−5t
2
Hence, the equation of linear and curved path is given by
x
2
−x
1
=15t (Linear path)
x
2
−x
1
=200+30t−5t
2
(Curved path)