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)
This question has multiple options
A. The velocity of projection of rings is \[\dfrac{{3g}}{4}\].
B. the maximum height attained by rings is \[\dfrac{g}{{32}}\].
C. When the first ring returns to her hand, the second ring is coming downwards and it is on the height of \[\dfrac{g}{4}\](from the ground).
D. When the first ring returns to her hand the third ring is going up and has travelled a distance of \[\dfrac{g}{4}\](from the ground).

Answer 1

Hint: In the solution we will be using the first equation of motion. First equation of motion gives us the relation between initial velocity, final velocity and time period of a body.

 Complete step by step answer:
It is given that the circus girl catches the first ring half second after the third is thrown.
\[\begin{array}{l}
{t_1} = \left( {\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2}} \right){\rm{ s}}\\
 = \dfrac{3}{2}{\rm{ s}}
\end{array}\]

First equation of motion for downward movement of a body under gravity.
\[v = u - gt\]......(1)

Here v is the final velocity, u is the initial velocity and t is the time period.

Third equation of motion for downward movement of a body under gravity.
\[s = ut - \dfrac{1}{2}gt{}^2\]……(2)

Here s is the distance travelled by the body.

Third equation of motion for downward movement of a body under gravity.
\[{v^2} = {u^2} - 2gs\]……(3)

It can be easily said that for half of the time \[{t_1}\] first ring goes upward and for the remaining time it comes downward.

It can also be understood that at maximum height the velocity of the first ring is zero.

Substitute \[0\] for \[v\] and \[\dfrac{3}{4}{\rm{ s}}\] for \[t\] in equation (1).
\[\begin{array}{l}
0 = u - g\left( {\dfrac{3}{4}{\rm{ s}}} \right)\\
u = \dfrac{{3g}}{4}
\end{array}\]

Here \[u\] is the velocity of projection of rings.

Substitute \[0\] for v and \[\left( {\dfrac{{3g}}{4}} \right)\] for u in equation (2).
\[\begin{array}{l}
{0^2} = {\left( {\dfrac{{3g}}{4}} \right)^2} - 2gs\\
s = \dfrac{{9g}}{{32}}
\end{array}\]

Here \[s\] is the maximum height attained by the rings.

When the first ring returns to her hand, the travel time of the second ring becomes one second. It can also be said that one-fourth part of this one second is taken by the second ring for coming downward.

Substitute \[\dfrac{1}{4}\] for t, \[0\] for u in equation (2).
\[\begin{array}{l}
s = 0 \cdot t - \dfrac{1}{2}g\left( {\dfrac{1}{4}} \right){}^2\\
 = \dfrac{g}{{32}}
\end{array}\]

Here s is the downward distance travelled by the second ring for one-fourth of one second time.

When first ring returns to her hand, the second ring was coming downwards and it is on the height of H which is given by:
\[\begin{array}{l}
H = \left( {{\text{maximum height}}} \right) - \left( {{\text{downward distance}}} \right)\\
 = \dfrac{{9g}}{{32}} - \dfrac{g}{{32}}\\
 = \dfrac{g}{4}
\end{array}\]

Time period of the third ring when first is received is given as \[\dfrac{1}{2}\] second.

Substitute \[\left( {\dfrac{1}{2}} \right)\] for t and \[\left( {\dfrac{{3g}}{4}} \right)\] for u in equation (2).

 \[\begin{array}{l}
s = \left( {\dfrac{{3g}}{4}} \right) \cdot \dfrac{1}{2} - \dfrac{1}{2}g{\left( {\dfrac{1}{2}} \right)^2}\\
 = \dfrac{g}{4}
\end{array}\]

So, the correct answer is “Option A,C,D”.

Note:
Read the given option’s statements carefully and analyse them correctly otherwise the solution would become incorrect. It is also to be taken care at what instant the ring is thrown and caught.