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IIT JEE Advanced 2011 Paper 2
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© examsnet.com
Question : 15
Total: 60
A train is moving along a straight line with a constant acceleration a. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 60° to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in
m
∕
s
2
, is _____.
Your Answer:
Validate
Solution:
We have
s = ut +
1
2
a
t
2
Therefore, the vertical displacement is
S
v
= 0 = u sin 60° t +
1
2
g
t
2
⇒
10
√
3
2
t
−
1
2
10
t
2
= 0
or t =
√
3
s. The horizontal displacement is
S
H
= 1.15 m = ut cos 60° -
1
2
a
t
2
⇒ 1.15 =
5
√
3
−
a
2
3
Therefore, a =
2
3
(
5
√
3
−
1.15
)
= 5
m
∕
s
2
.
© examsnet.com
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