The correct option is C 30 ms−1
Given,
initial velocity, u=10 ms−1
time taken, t = 5 s
distance covered, s = 100 m
Let the final velocity be 'v' and acceleration be 'a'.
From the second equation of motion,
s=ut+12×at2,
100=10×5+12×a×52
∴a=4 ms−2
From the first equation of motion,
v = u + at,
v=10+4×5 = 30 ms−1
Therefore the final velocity is 30 ms−1.