Question

A balloon is rising with constant acceleration $2m/{\sec }^2$.Two stones are released from the balloon at the interval of $2$ sec. Find out the distance between the two stones $1$ sec after the release of second stone.

Answer 1

let A and B be 1st and 2nd stone to be released

interval $\left(1\right)$ : 2 sec after release of first stone$\left(A\right)$

let $s_1$ be the sepration between A and B at the end of this interval

$u_{ba}=0$

$a_{ba}=2-(-10)=12$

$s_1=u_{ba}\times 2+\frac{a_{ba}\times 2^2}{2}$

$s_1=24m\dots \left(a\right)$

$v_{ba}=a_{ba}\times 2$

$v_{ba}=24m/s\dots \left(1\right)$

interval $\left(2\right)$ : 1 sec after release of 2nd stone$\left(B\right)$

let $s_2$ be the sepration increased between A and B at the end of this interval

at $t=0$ of this interval

$u_{ba}=v_{ba}$ of interval $\left(1\right)$

$u_{ba}=24m/s$

$a_{ba}=10-10=0$

$s_2=u_{ba}\times 1+\frac{0\times 1^2}{2}$

$s_2=24m\dots \left(b\right)$

final sepration between A and B $=s_1+s_2$

$=24+24$

$=48m$