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 between the two bodies two second after the release of the second body is

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 between the two bodies two second after the release of the second body is (A) 9.8 m (B) 4.9 m (C) 24.5 m (D) 19.6 m

Tardigrade Admin · Accepted Answer

As, Δ x =(1/2) g t2-(1/2) g(t-1)2 =(1/2) g[-(t-1)2]=(1/2) g(2 t-1) =(1/2) × 9.8 × 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 between the two bodies two second after the release of the second body is

9.8 m

4.9 m

24.5 m

19.6 m

As, $Δ x = \frac{1}{2} g t^{2} - \frac{1}{2} g (t - 1)^{2}$
$= \frac{1}{2} g [- (t - 1)^{2}] = \frac{1}{2} g (2 t - 1)$
$= \frac{1}{2} \times 9.8 \times 5 m = 24.5 m$

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 between the two bodies two second after the release of the second body is

9.8 m

4.9 m

24.5 m

19.6 m

As, Δx=21​gt2−21​g(t−1)2 =21​g[−(t−1)2]=21​g(2t−1) =21​×9.8×5m=24.5m

As, $Δ x = \frac{1}{2} g t^{2} - \frac{1}{2} g (t - 1)^{2}$
$= \frac{1}{2} g [- (t - 1)^{2}] = \frac{1}{2} g (2 t - 1)$
$= \frac{1}{2} \times 9.8 \times 5 m = 24.5 m$