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Question

A circus girl throws three rings upwards one after the other at equal intervals of half a second. She catches the first ring half second after the third was thrown. Then,
(g= acceleration due to gravity)

A
the velocity of projection of rings is 3g4
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B
the maximum height attained by rings is g32
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C
when the first ring returns to her hand, the second ring was coming downwards and it is on the height of g4 (from the ground).
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D
when the first ring returns to her hand, the third ring was going up and has travelled a distance of g4 (from the ground).
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Solution

The correct options are
A the velocity of projection of rings is 3g4
C when the first ring returns to her hand, the second ring was coming downwards and it is on the height of g4 (from the ground).
D when the first ring returns to her hand, the third ring was going up and has travelled a distance of g4 (from the ground).
The circus girl catches the first ring after 12+12+12=32 second
the ring was in air for32 second in which for first 34 second it was going up and for the next 34 second it was coming down.
Let the height attained by first ring be h in 34 second, obviously, its velocity at the instant will be zero.
if u be the velocity of projection of rings, then
0=ug(34)u=3g4
(A) is correct.
The maximum height h attained by the ring is given by,0=(3g4)22gh
h=9g32(B) is not correct.
When the first ring returns to her hand, the second ring has traveled for 1 second. Of that 1 second, 34 seconds will be spent in going up and for the rest it is coming down.
h=12g(14)2=g32
Distance travelled by second ring by that instant is=9g32g32=g4
(C) is correct.
The third ring has travelled for t=12 second by that instant. it will be going up.
Corresponding distance from groundt=(3g4)(12)12g(12)2
=3g8g8=g4
(D) is correct.

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