Question

A juggler throws up balls at regular intervals of time. Each ball takes $2s$ to reach the highest position. If the first ball is in the highest position by the time the fifth one starts, then the separation between the first and the second balls is

• A. 1.225m
• B. 2.45m
• C. 4.9m
• D. 3.8m

Answer 1

The correct option is A 1.225m

Let the interval between each balls be $\Delta t$,
$\Rightarrow$ difference between $1^{st}$ and $5^{th}$ ball $=4\Delta t$
$\Rightarrow 4\Delta t=2\Rightarrow \Delta t=\frac{1}{2}:Also,4\Delta t=\frac{u}{g}$
$u=4\Delta tg-\left(1\right)$

So, separation between $1^{st}$ and $2^{nd}$ ball,
$s_1=u\left(2\right)-\frac{1}{2}g(2{)}^2$
$s_2=u(2-\Delta t)-\frac{1}{2}g(2-\Delta t{)}^2$

So, $s_1-s_2=u\left(\Delta t\right)-\frac{1}{2}g(4\Delta t-\Delta t^2)$
From (1) $=4(\Delta t{)}^{2}g-\frac{g}{2}(4\Delta t-\Delta t^2)$
$=4(\frac{1}{2}{)}^{2}g-\frac{g}{2}(\frac{4}{2}-\frac{1}{4})=g-\frac{g}{2}(\frac{7}{4})$
$=1.225mt$