A small fish is dropped by a pelican that is rising steadily at 0.65 m/s. (Assume the positive direction is upward.)

1. After 2.5 s, what is the velocity of the fish?
2. How far below the pelican is the fish after 2.5 s?

30.6 meters below the pelican

To solve the problem, we can use the equations of motion. Let's assume that the initial velocity of the fish is zero.

1. After 2.5 s, what is the velocity of the fish?

To find the velocity of the fish, we can use the equation:

Final velocity (v) = Initial velocity (u) + (acceleration) * (time)

Here, the acceleration is the acceleration due to gravity, which is approximately -9.8 m/s^2 (since it acts in the opposite direction to the positive direction).

Plugging in the values:
u = 0 m/s (initial velocity)
a = -9.8 m/s^2 (acceleration)
t = 2.5 s (time)

v = 0 + (-9.8) * 2.5
v = -24.5 m/s

Therefore, after 2.5 seconds, the velocity of the fish is -24.5 m/s.

2. How far below the pelican is the fish after 2.5 s?

To find the distance traveled by the fish, we can use the equation:

Distance (s) = Initial velocity (u) * (time) + (0.5) * (acceleration) * (time)^2

Plugging in the values:
u = 0 m/s (initial velocity)
a = -9.8 m/s^2 (acceleration)
t = 2.5 s (time)

s = 0 * 2.5 + 0.5 * (-9.8) * (2.5)^2
s = -30.625 m

Therefore, after 2.5 seconds, the fish is approximately 30.625 meters below the pelican.

To solve these problems, we can use the formulas of kinematics. Kinematics is the branch of physics that deals with the motion of objects.

1. To find the velocity of the fish after 2.5 seconds, we can use the equation of motion:

v = u + at

where:
v = velocity of the fish after 2.5 seconds
u = initial velocity of the fish (which is 0 m/s because the fish is dropped and initially at rest)
a = acceleration (which is the gravitational acceleration, approximately 9.8 m/s^2)
t = time (which is 2.5 seconds)

Plugging in these values, we get:

v = 0 + (9.8)(2.5)
v = 0 + 24.5
v = 24.5 m/s

Therefore, the velocity of the fish after 2.5 seconds is 24.5 m/s upwards.

2. To find how far below the pelican the fish is after 2.5 seconds, we can use another equation of motion:

s = ut + (1/2)at^2

where:
s = distance traveled by the fish after 2.5 seconds
u = initial velocity of the fish (which is 0 m/s because the fish is dropped and initially at rest)
a = acceleration (which is the gravitational acceleration, approximately 9.8 m/s^2)
t = time (which is 2.5 seconds)

Plugging in these values, we get:

s = (0)(2.5) + (1/2)(9.8)(2.5)^2
s = 0 + (1/2)(9.8)(6.25)
s = 0 + 30.625
s = 30.625 m

Therefore, after 2.5 seconds, the fish is approximately 30.625 meters below the pelican.