Question

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

Answer 1

Displacement of the first body after 3 seconds= S = ut + ½ g t^2  = 44.1m/s

Displacement of the first body after 2 seconds= S = ut + ut + ½ g t^2  = 19.6m/s

The separation between them=44.5 - 19.6 = 24.5m/s

Answer 2

Answer:

Explanation:we shall sue the following formula

s = ut + (1/2)at2

here u = 0 m/s and g = 10 m/s2

now,

for te first body the duration it has travelled will be 1s + 2s = 3s

for the second body the duration it has travelled will be = 2s

everything else is same for both cases

so,

for 1st

s = (1/2)gt2 = (1/2) x 10 x 9 = 45m

for 2nd

s' = (1/2)gt'2 = (1/2) x 10 x 4 = 20m

thus, the separation between te two bodies,two second after the release of second body will be

s - s' = 45m - 20m = 25 m

which is option(4).