Question

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Answer 1

Answer:

A) 24.5 m

Explanation:

Given,

• The body is released, which means initial velocity u = 0
• Another body is released after one second.

To find, Separation distance of both the bodies 2 sec after the release of second body is nothing but difference of distance travelled by first body in first 3 seconds and the distance travelled by second body in 2 seconds after it's drop.

From equations of motion for uniformly accelerated body:

s = ut + at^2/2

s = at^2/2 ------(1)

Separation distance = s1 - s2

= at1^2/2 - at2^2/2

= a/2 ( t1^2 - t2^2)

= g/2( 3^2 - 2^2)

= 9.8/2 (5)

= 4.9(5)

= 24.5 m

Hence, separation between both bodies two seconds after the release of second body = 24.5m