Question

A body is released from a height and it falls freely towards the earth. Exactly 1 s later another body is released. What is the distance between the two bodies 2 s after the release of the second body?

• A. 4.9 m
• B. 9.8 m
• C. 24.5 m
• D. 50 m

Answer 1

Answer: (C)

24.5 m

Initial speed of the bodies $u=0$.

Distance covered by the first body in $t=3$ s, $S_1=ut+\frac{1}{2}g{t}^2$

Or $S_1=0+\frac{1}{2}\times 9.8\times 3^2=44.1$ m

Distance covered by the second body in $t^{\prime }=2$ s, $S_2=u{t}^{\prime }+\frac{1}{2}g(t^{\prime }{)}^2$

Or $S_2=0+\frac{1}{2}\times 9.8\times 2^2=19.6$ m

Thus distance between the bodies $\Delta S=S_1-S_2$

$\therefore$ $\Delta S=44.1-19.6=24.5$ m