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.
Pingback:Numerical Problem 2, Forces and Motion … msa – msa
Pingback:Numerical Problems on Forces and Motion – msa