Question

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

• A. $48m$
• B. $84m$
• C. $40m$
• D. $60m$

Answer 1

The correct option is A $48m$

Let the velocity of the balloon be $u_1$ and time instant be $t=0$ when the first stone is released.

Motion of first stone in 3 sec,

$s_1=3u_1-\left(1/2\right)\times g\times (3{)}^2$

$=3u_1-9g/\mathrm{2............}\left(i\right)$

Motion of second stone in first two seconds is same as motion of balloon,

$u_2=u_1+2\times 2$

$=u_1+\mathrm{4................}\left(ii\right)$

$s_{2_{0to2}}=2u_1+\left(1/2\right)\times 2\times 2^2$

$=2u_1+\mathrm{4...............}\left(iii\right)$

Motion of second stone in third second

$s_{2_{2to3}}=u_2\left(1\right)-\left(1/2\right)g(1{)}^2$

$=u_2-g/\mathrm{2...............}\left(iv\right)$

Substituting $\left(ii\right)$ in $\left(iv\right)$,

$s_{2_{2to3}}=u_1+4-g/\mathrm{2..........}\left(v\right)$

From $\left(iii\right)\&\left(v\right),$ Displacement of second particle in 3 s,

$s_2=s_{2_{0to2}}+s_{2_{2to3}}$

$=3u_1+8-g/2$

Distance between the two stones after 3 sec,

$s_2-s_1=3u_1+8-g/2-(3u_1-9g/2)$

$=8+4g$

$=48m$