I think it's 39.53
(please do calculate it. I am not completely sure)
The distance to the water is the same for both ... call it s meters
They both take 1.6 s to reach the water ... so t = 1.6 seconds
They both step off the bridge ... so both have an initial vertical velocity is 0 m/s
Just consider the vertical motion and take DOWN as the positive direction
a = 9.8 m/s²
s = v(i)t + (1/2)at²
s = 0 + (1/2) * 9.8 * 1.6²
s = 12.5 m ←←← Edit: Forgot to say ... that's how high the bridge is above the water
Now get the time it takes the first jumper to reach 1.8 m:
s = v(i)t + (1/2)at²
1.8 = 0 + 4.9t²
t = 0.61 s
so when the 2nd person jumps it takes the first person another 1.6 - 0.61 = 0.99 s to reach the water
Now find how far the 2nd jumper falls in 0.99 s:
s = 0 + 4.9 * 0.99²
s = 4.8 m
so the separation distance between the two jumpers when the 1st jumper hits the water is 12.5 - 4.8 = 7.7 m
