A ball is dropped from a bridge 122.5m above a river. After the ball has been falling for 2s, a second ball is thrown straight down after it. What must the initial velocity of the second ball be so that both hit the water at the same time?

Correct option is C. 26.1m/s

Given that,
Height h = 122.5 m
Time t = 2 sec

Time taken by the first ball to hit water can be found as:
Using equation of motion, s = ut + (1/2)gt²
122.5 = 0 + (1/2) × 9.8 × t²
t² = 2 × 122.5 / 9.8
t² = 25
t = 5 sec

Now the next ball must cover this distance in 3 s (5s - 2s) to hit water at the same time.
122.5 = 3u + (1/2) × 9.8 × 3²
122.5 = 3u + 44.1
3u = 78.4
u = 26.1 m/s

So, the initial velocity of the ball is 26.1 m/s
Hence, C is the correct option.