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?

Your solution is missing the additional distance that the pelican traveled in the 2.8 s.

To calculate the velocity of the fish after 2.8 s, we need to consider the initial velocity and the acceleration due to gravity.

Given:
Initial velocity of the fish = 0 m/s (since it was dropped)
Acceleration due to gravity = 9.81 m/s^2 (downward)

To calculate the final velocity of the fish, we can use the equation:
v = u + at

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time taken

Substituting the values into the equation, we have:
v = 0 + (9.81)(2.8)

Calculating the value, we get:
v = 0 + 27.498
v ≈ 27.5 m/s

Therefore, the velocity of the fish after 2.8 s is approximately 27.5 m/s.

Now let's calculate the distance below the pelican the fish is after 2.8 s using the formula for distance traveled under constant acceleration:

s = ut + 0.5at^2

Given:
u = 0 m/s (initial velocity)
a = 9.81 m/s^2 (acceleration due to gravity)
t = 2.8 s (time taken)

Substituting the values into the formula, we have:
s = 0 + 0.5(9.81)(2.8^2)

Calculating the value, we get:
s ≈ 37.15 m

Therefore, the fish is approximately 37.15 meters below the pelican after 2.8 s.

To find the velocity of the fish after 2.8 seconds, we can use the equation of motion:

vf = vi + at

where:
vf = final velocity
vi = initial velocity
a = acceleration
t = time

In this case, the pelican is rising steadily at a velocity of 0.60 m/s, so the initial velocity (vi) of the fish is 0.60 m/s. The acceleration (a) is the acceleration due to gravity, which is 9.81 m/s². The time (t) is given as 2.8 seconds.

Plugging in the values into the equation, we get:

vf = 0.60 m/s + (9.81 m/s²)(2.8 s)
vf = 0.60 m/s + 27.42 m/s
vf = 27.42 m/s

Therefore, the velocity of the fish after 2.8 seconds is 27.42 m/s.

To find the distance below the pelican that the fish is after 2.8 seconds, we can use another equation:

d = vi*t + (1/2)at²

where:
d = distance
vi = initial velocity
t = time
a = acceleration

Since the fish is dropped, its initial velocity (vi) is 0 m/s. The acceleration (a) is still the acceleration due to gravity, which is 9.81 m/s². The time (t) is given as 2.8 seconds.

Plugging in the values into the equation, we get:

d = (0 m/s)(2.8 s) + (1/2)(9.81 m/s²)(2.8 s)²
d = 0 m + (1/2)(9.81 m/s²)(7.84 s²)
d = 0 m + (1/2)(77.1224 m)
d = 0 m + 38.5612 m
d = 38.5612 m

Therefore, the fish is 38.5612 meters below the pelican after 2.8 seconds.

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.