1.

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 mB. 9.8 mC. 19.6 mD. 1 m

Answer»

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.


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