Question

A body released from a great height falls freely towards the earth. Another body is released from the same height exactly one second later. The separation (in $m$ ) between the two bodies two seconds after the release of the second body is ( $g=9.8 \, m \, s^{- 2}$ )

Answer 1

Separation of the body after $t$ second after the release of 1st body is
$\Delta x=\frac{1}{2}gt^{2}-\frac{1}{2}g\left(t - 1\right)^{2}$
$=\frac{1}{2}g\left[t^{2} - \left(t - 1\right)^{2}\right]=\frac{1}{2}g\left(2 t - 1\right)$ $t=2+1=3$
$\Delta x=\frac{1}{2}\times 9.8\times 5m=24.5m$