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 $\frac{3g}{4}$
• B. the maximum height attained by rings is $\frac{g}{32}$
• C. when the first ring returns to her hand, the second ring was coming downwards and it is on the height of $\frac{g}{4}$ (from the ground).
• D. when the first ring returns to her hand, the third ring was going up and has travelled a distance of $\frac{g}{4}$ (from the ground).

Answer 1

Answer: (A), (C), (D)

The correct options are
A the velocity of projection of rings is $\frac{3g}{4}$
C when the first ring returns to her hand, the second ring was coming downwards and it is on the height of $\frac{g}{4}$ (from the ground).
D when the first ring returns to her hand, the third ring was going up and has travelled a distance of $\frac{g}{4}$ (from the ground).

The circus girl catches the first ring after $\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}$ second
$\Rightarrow$ the ring was in air for$\frac{3}{2}$ second in which for first $\frac{3}{4}$ second it was going up and for the next $\frac{3}{4}$ second it was coming down.
Let the height attained by first ring be h in $\frac{3}{4}$ second, obviously, its velocity at the instant will be zero.
if u be the velocity of projection of rings, then
$0=u-g\left(\frac{3}{4}\right)\Rightarrow u=\frac{3g}{4}$
$\Rightarrow \left(A\right)$ is correct.
The maximum height h attained by the ring is given by,$0={\left(\frac{3g}{4}\right)}^2-2gh$
$\Rightarrow h=\frac{9g}{32}\Rightarrow \left(B\right)$ is not correct.
When the first ring returns to her hand, the second ring has traveled for 1 second. Of that 1 second, $\frac{3}{4}$ seconds will be spent in going up and for the rest it is coming down.
$\Rightarrow h^{\prime }=\frac{1}{2}g{\left(\frac{1}{4}\right)}^2=\frac{g}{32}$
$\Rightarrow$ Distance travelled by second ring by that instant is$=\frac{9g}{32}-\frac{g}{32}=\frac{g}{4}$
$\Rightarrow$ (C) is correct.
The third ring has travelled for $t=\frac{1}{2}$ second by that instant. it will be going up.
Corresponding distance from ground$t=\left(\frac{3g}{4}\right)\left(\frac{1}{2}\right)-\frac{1}{2}g{\left(\frac{1}{2}\right)}^2$
$=\frac{3g}{8}-\frac{g}{8}=\frac{g}{4}$
$\Rightarrow \left(D\right)$ is correct.