Question

A body released from a height falls eely towards
the earth. Another body is released rom the same
height exactly a second later. Then the separation
between the two bodies two second after the
release of the second body is

Answer 1

Explanation:

from law to kinematics

s=ut+1/2at^2

u =0

a=g as it is freely falling

for the first body time is 3sec( 1+2)

s=1/2gt^2

s1=1/2×10×9

s1=45m

for second body t=2sec

s=1/2×10×4

s2=20m

seperation= s1-s2

=25m

Answer 2

$the \: second \: ball \: is \: released \: a \: second \: \: later \: of \: first \: ball \: \:$

SO WE CAN SAY THAT THE TIME OF FLIGHT OF FIRST BALL IS 1 SEC

USING 3RD EQN.

$s = ut \: + \frac{1}{2} g {t}^{2} \\ \\ here \: u \: is \: zero \: \\ \\ so \: s = \frac{1}{2} 10 \times 1 = 5m$

so distance between them is 5 m

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