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AIPMT 2010 Physics and Chemistry Paper
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© examsnet.com
Question : 8
Total: 100
A ball is dropped from a high platform at
t
=
0
starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed
v
. The two balls meet at
t
=
18
s
. What is the value of
v
? (Take
g
=
10
m
/
s
2
)
75
m
/
s
55
m
/
s
40
m
/
s
60
m
/
s
Validate
Solution:
Let the two balls meet after
t
s
at distance
x
from the platform.
For the first ball
u
=
0
,
t
=
18
s
,
g
=
10
m
/
s
2
Using
h
=
u
t
+
1
2
g
t
2
∴
x
=
1
2
g
t
2
2
....(i)
u
×
12
+
1
2
×
10
×
(
12
)
2
For the second ball
u
=
u
,
t
=
12
s
,
g
=
10
m
/
s
2
Using
h
=
u
t
+
1
2
×
10
×
10
2
....(ii)
From equations (i) and (ii), we get
1
2
×
10
×
18
2
=
12
u
+
1
2
×
10
×
(
12
)
2
or
12
u
=
1
2
×
10
×
[
(
18
)
2
−
(
12
)
2
]
=
1
2
×
10
×
[
(
18
+
12
)
(
18
−
12
)
]
12
u
=
1
2
×
10
×
30
×
6
or
u
=
1
×
10
×
30
×
6
2
×
12
=
75
m
/
s
© examsnet.com
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