Question

A body released from a height falls freely towards earth. Another body is released from the same height exactly one second later. The seperation between the two bodies two second after the release of the second body is:

Answer 1

Answer:

4.9 meters

$s = ut + \frac{1}{2} at^{2}$

u = initial speed = 0

t = 1 s

a = 9.8m/s²

$s = 0 \times 1 + \frac{1}{2} \times 9.8 \times 1^{2}$

$s = \frac{1}{2} \times 9.8$

$s = 4.9$

The separation is of 4.9 m

Answer 2

Answer:

$(d) \: 24.5m$

Explanation:

$time \: for \: {1}^{st} \: ball = 2 \: seconds$

$time \: for \: {2}^{nd} \: ball = 2 + 1 = 3 \: seconds$

$S_{1} = \frac{1}{2} g( {2}^{2} ) \: and \: S_{2} = \frac{1}{2} g( {3}^{2} )$

$S_{2} - S_{1} = g / 2(9 - 4) = 4.9 \times 5 = 24.5m$

$hence \: \: verified$