V=Vo + g*t = -0.60 + 9.81*2.8=26.87 m/s
d = Vo*t + 0.5g*t^2
d = -0.60*2.8 + 4.9*2.8^2 = 36.74 m below the pelican.
A small fish is dropped by a pelican that is
rising steadily at 0.60 m/s.
After 2.8 s, what is the velocity of the fish?
The acceleration of gravity is 9.81 m/s2.
How far below the pelican is the fish after the
2.8 s?
2 answers
Your solution is missing the additional distance that the pelican traveled in the 2.8 s.