Question

An object is released from some height. Exactly after one second, another object is released from the same height. The distance between the two objects exactly after $2$ seconds of the release of second object will be

• A. $4.9\text{m}$
• B. $9.8\text{m}$
• C. $19.6\text{m}$
• D. $24.5\text{m}$

Answer 1

The correct option is D $24.5\text{m}$

If time $t=2\text{sec}$ according to second object then total time elapsed by 1st particle $=3\text{sec}$
$\therefore$ Displacement of the respective particles are given by the formula
$S=ut+\frac{1}{2}a{t}^2=\frac{1}{2}a{t}^2$ [As $u=0$]
Thus, we have
$S_1=\frac{1}{2}\times 9.8\times 3^2$
and, $S_2=\frac{1}{2}\times 9.8\times 2^2$
Hence, distance between two objects is
$S_1-S_2=\frac{1}{2}\times 9.8(3^2-2^2)=24.5\text{m}$