if the pelican in item 3 was traveling at the same speed but was only 2.7m above the water, how far would the fish travel horizontally before hitting the water?​

5.66 m

Explanation:

From online sources, the speed in item 3 being referred to was discovered to be 7.62 m/s

Now, let's get the time of flight from one of Newton's equation of motion;

S = ut + ½gt²

Considering the vertical component, we have u = 0 m/s.

Thus;.

S = ½gt²

Plugging in the relevant values;

2.7 = ½ × 9.8 × t²

t² = 2.7/4.9

t = √(2.7/4.9)

t = 0.7423 s

Now, when we consider the horizontal component of the motion, we have;

S = vt

Where;

S is the distance the fish will travel horizontally before hitting the water.

v = 7.62 m/s

t = 0.7423

Thus

s = 7.62 × 0.7423

s ≈ 5.66 m