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Byju's Answer
Standard IX
Physics
Derivation of Position-Velocity Relation by Graphical Method
the relative ...
Question
the relative velocity of the ball with respect to the elevator when the ball hits the elevator.
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Solution
For ball :
u
=
18
m
/
s
a
=
−
g
=
−
9.8
m
/
s
2
Velocity of elevator
v
=
2
m
/
s
Time taken by the ball to strike the elevator
t
=
3.65
s (solved earlier)
For ball :
V
y
=
u
−
g
t
∴
V
y
=
18
−
9.8
×
3.65
=
−
17.77
m
/
s
Relative velocity of ball w.r.t elevator
V
b
e
=
V
y
−
v
=
−
17.77
−
2
=
−
19.77
m
/
s
∴
V
b
e
=
19.77
m
/
s
(downwards)
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Similar questions
Q.
A ball is thrown vertically upward from the 12 m level with an initial velocity of 18 m/s. At the same instant an open platform elevator passes the 5 m level, moving upward with a constant velocity of 2 m/s. Determine (g = 9.8
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(a) when and where the ball will meet the elevator
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Q.
A man in a lift ascending with an upward acceleration throws a ball vertically upwards and catches it after
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A) The acceleration of the elevator is
g
(
t
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−
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+
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)
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+
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.
We can conclude that:
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