Ip standing side by side, you and a friend step off a bridge at different times and fall for 1.6 s to the water below. your friend goes first, and you follow after she has dropped a distance of 1.8 m .

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