## Problem 1: An object is falling freely under the gravity. How much distance will it travel in 2nd and 3rd seconds of its journey?

##### Explanation

Consider the figure. The ball at position A starts falling with initial velocity 0 m/s. After the first second, it reaches position B attaining velocity Vf1. This velocity is now the initial velocity for its journey in the 2nd second. At the end of 2nd second, it reaches position C traversing distance S2 and attaining velocity vf2 which is now initial velocity vi3 for its 3rd second journey. After 3rd second, it reaches position C with velocity vf3. Now we calculate these distances. Solution

For the first second,

Vi = 0 m/s, t = 1 s, vf = ?

Now, vf = vi + gt ⇒ vf = 0 + 9.8 = 9.8 m/s

This velocity serves as initial velocity for the 2nd second.

For the 2nd second

Vi = 9.8 m/s, t = 1 s, S = ?

Apply S = vtt + ½ gt2, we have,

S2 = 9.8 × 1 + ½ × 9.8 × 12 = 9.8 + 4.9 = 14.7 m

This is the distance covered in 2nd second.

Now the velocity at the end of 2nd second can be calculated as vf = vi + gt. Put the values,

Vf = 9.8 + 9.8 × 1 = 19.6 m/s and this is the initial velocity at the start of 3rd second.

For the 3rd second

Vi = 19.6, t = 1 s, g = 9.8 m/s2, S = ?

Apply the second equation of motion once again,

S3 = 19.6 × 1 + ½ × 9.8 × 12 = 19.6 + 4.9 = 24.5 m

This is the distance covered in 3rd second.