Question

A balloon is ascending vertically with an acceleration of 0.2 m $s^{-2}$ . Two stones are dropper from it at an interval of 2 s. The distance between them when the second stone is dropped is (take g = 9.8 m $s^{-2}$):

• A. 24 m
• B. 4.9 m
• C. 19.6 m
• D. 20.0 m

Answer 1

The correct option is A 24 m

A balloon is ascending vertically with an acceleration= $0.2m/s^2$

Two stones are dropped from it at an interval of $t=2s$ .

$g=9.5m/S$

The distance between them when the $2nd$ stone is dropped:

$s_1=\frac{1}{2}a{t}^2$ $(\because u=0)$

or, $s_1=\frac{1}{2}\times 0.2\times 2^2$

$=\frac{1}{2}\times \frac{2}{10}\times 2^2$

$=\frac{4}{10}=0.4m$ .

Due to free fall the $1st$ stone will have,

$s=\frac{1}{2}\times (0.2+9.8)\times 2^2$

$=\frac{1}{2}\times 10\times 4$

$=20$ .

$\therefore$ Hence the $2nd$ stone is dropped= $(20+4)=24m$.