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 )

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 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Separation of the body after t second after the release of 1st body is Δ x=(1/2)gt2-(1/2)g(t - 1)2 =(1/2)g[t2 - (t - 1)2]=(1/2)g(2 t - 1) t=2+1=3 Δ x=(1/2)× 9.8× 5m=24.5m

Q. 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 ( $g = 9.8 m s^{- 2}$ )

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

Q. 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=9.8ms−2 )

Separation of the body after t second after the release of 1st body is Δx=21​gt2−21​g(t−1)2 =21​g[t2−(t−1)2]=21​g(2t−1) t=2+1=3 Δx=21​×9.8×5m=24.5m

Q. 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 ( $g = 9.8 m s^{- 2}$ )

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