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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A man holds four balls 180 m above the ground and drops them at regular intervals of time so that when the first ball hits ground, fourth ball is just leaving his hand. At this time the second and third balls from the ground are at the positions.

a)160 m and 100 m

b)80 m and 20 m

c)20 m and 80 m

d)100 m and 160 m

Can you answer this question?

1 Answer

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Time taken by first ball h reach the ground
$=\sqrt {\large\frac{2H}{g}}=\sqrt {\large\frac{2 \times 180}{10}}$$=6 sec$
time interval between two balls in 2 seconds
$H_1=\large\frac{1}{2} $$g(4)^2=80 m$
$H_1=\large\frac{1}{2} $$g(2)^2=20 m$
Height from ground are $180-80=100m$
$180 -20 =160 m$
Hence d is the correct answer.


answered Jul 3, 2013 by meena.p
edited Jan 25, 2014 by meena.p
Meena ji..i'm confused..where is the value of u in the equations of H1 and H2..? Pls reply.

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