A circus girl throws up three rings one after the other at equal intervals of half second. She catches the first ring half second after the third ring was thrown. Then, A. the velocity of projection of the rings B. he height of the second ring when the first ring returns to her(in mm) 1. 0.75 2. 2 3. 2.5 4. 7.5 5. 10
Answers
Answer:
Type your question...
11th
Physics
Motion in a Straight Line
Problems on Equation of Motion
A circus girl throws three ...
PHYSICS
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)
This question has multiple correct options
December 30, 2019
avatar
Keerthi Dsouza
Share
Save
ANSWER
The circus girl catches the first ring after
2
1
+
2
1
+
2
1
=
2
3
second
⇒ the ring was in air for
2
3
second in which for first
4
3
second it was going up and for the next
4
3
second it was coming down.
Let the height attained by first ring be h in
4
3
second, obviously, its velocity at the instant will be zero.
if u be the velocity of projection of rings, then
0=u−g(
4
3
)⇒u=
4
3g
⇒(A) is correct.
The maximum height h attained by the ring is given by,0=(
4
3g
)
2
−2gh
⇒h=
32
9g
⇒(B) is not correct.
When the first ring returns to her hand, the second ring has traveled for 1 second. Of that 1 second,
4
3
seconds will be spent in going up and for the rest it is coming down.
⇒h = 32/g
⇒ Distance travelled by second ring by that instant is=
32/9g - 32/g = 4/g
⇒ (C) is correct.
The third ring has travelled for t=
2
1
second by that instant. it will be going up.
Corresponding distance from ground =
= 3g/8 − g/8 = g/4