Question

Two bodies, one held 1 m vertically above the other, are released simultaneously and fall freely under gravity. After 2 second, the relative separation of the bodies will be:

A. 4.9 m

B. 9.8 m

C. 19.6 m

D. 1 m

Answer 1

Now, initial separation = 1 m

Body 1:

Initial Speed = 0

Acceleration = 9.8 ms^-2

Time = 2 seconds

Now, we know that,

S = ut + \(\frac{1}{2}\)at²

So, S₁ = \(\frac{1}{2}\)x 9.8 x 4

S₁ = 19.6 m

Body 2:

Initial Speed = 0

Acceleration = 9.8 ms^-2

Time = 2 seconds

Now, we know that,

S = ut + \(\frac{1}{2}\)at²

^So, S₂ ⁼ \(\frac{1}{2}\)x 9.8 x 4

S₂ = 19.6 m

Since,S₁ = S₂ so final separation is also 1 m.

Hence, option D is correct.