A man in a lift ascending with an upward acceleration a throws a ball vertically upwards with a velocity v with respect to himself and catches it after t_1 seconds. After wards when the lift is descending with the same acceleration a acting downwards the man again throws the ball vertically upwards with the same velocity with respect to him and catches it after t_2 seconds?
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2
Answer:
We know that
t
1
=
g+a
2u
&t
2
=
g−a
2u
We get,
2u=t
1
(g+a)=t
2
(g−a)
a(t
1
+t
2
)=g(t
2
−t
2
)
a=
t
1
+t
2
g(t
2
−t
1
)
Now, the velocity of projection of the ball relative to elevator is u, We have,
g+a=
t
1
2u
g−a=
t
2
2u
adding both,
2g=2u(
t
1
1
+
t
1
1
)
∴u=
t
1
+t
2
t
1
t
2
g
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