**For the 2**^{nd} second

V_{i} = 9.8 m/s, t = 1 s, S = ?

Apply S = v_{t}t + ½ gt^{2}, we have,

S_{2} = 9.8 × 1 + ½ × 9.8 × 1^{2} = 9.8 + 4.9 = **14.7 m**

This is the distance covered in 2^{nd} second.

Now the velocity at the end of 2^{nd} second can be calculated as v_{f} = v_{i} + gt. Put the values,

V_{f} = 9.8 + 9.8 × 1 = 19.6 m/s and this is the initial velocity at the start of 3^{rd} second.

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