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RRB Paramedical Staff Nurse Model Paper 3
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© examsnet.com
Question : 94
Total: 100
A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact is
2100
m
s
e
c
2
downwards
2100
m
s
e
c
2
upwards
1400
m
s
e
c
2
700
m
s
e
c
2
Validate
Solution:
Velocity at the time of striking the floor,
u
=
√
2
g
h
=
√
2
×
9.8
×
10
=
14
m
/
s
(-ve since downwards)
Velocity with which it rebounds.
v
=
√
2
g
h
2
=
√
2
×
9.8
×
2.5
=
7
m
/
s
(+ve since upwards)
∴
Change in velocity
△
v
=
7
−
(
−
14
)
=
21
m
/
s
∴
Acceleration
=
∆
v
∆
t
=
21
0.01
=
2100
m
/
s
2
(upwards since positive)
© examsnet.com
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