nu ieziell all resistance).
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 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 = 10 m/s. Give the equations for the linear and curved
parts of the plot.
8
10
)
Answers
my equation answer is not coming
hope i help ____________ " - "
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)