Question

A balloon is ascending vertically with an acceleration of $0.2m/s^2$. Two stones are dropped from it at an interval of 2 sec. The distance (in m) between them 1.5 sec after the second stone is released is approximately 10x. The value of x is: (use $g=9.8m/s^2$)

Answer 1

The first stone travels for $2+1.5=3.5secs.$
The second stone travels for 1.5 secs.
$\text{Let first stone be dropped at t}=0$
$\text{Thus, distance travelled by it in 3.5 seconds is}$
$\frac{1}{2}g{t}^2=4.9(3.5{)}^2=60.025\text{m}$
$\text{Distance travelled by balloon upwards in 2 seconds}$
$=\frac{1}{2}a{t}^2=0.5\times 0.2\times 2^2=0.4\text{m}$
$\text{Distance travelled by stone 2 after being released}$
$=0.5\times 9.8\times {1.5}^2=11.025\text{m}$
$\text{Thus, distance between the 2 stones at required time}$
$=60.025-(-0.4+11.025)=49.4\text{m}$