2 stone's are up simultaneously from the edge of a cliff 200m high with intial speed 30m/s and 40m/s. draw the graph of varistion of the relative position of stone second with respect to stone second . also write essential expressions.
Answers
Answer:
it's a long one
Explanation:
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)