A small fish is dropped by a pelican that is
rising steadily at 0.44 m/s.
After 2.6 s, what is the velocity of the fish?
The acceleration of gravity is 9.81 m/s^2
Answer in units of m/s
& How far below the pelican is the fish after the
2.6 s?
Answer in units of m
To find the velocity of the fish after 2.6 s, we can use the equation of motion:
v = u + at
Where:
v is the final velocity of the fish
u is the initial velocity of the fish (which is zero since the fish was dropped)
a is the acceleration of the fish due to gravity (-9.81 m/s^2 since gravity pulls objects downward)
t is the time (2.6 s)
Plugging in the values, the equation becomes:
v = 0 + (-9.81) * 2.6
v = -25.506 m/s
Therefore, the velocity of the fish after 2.6 s is approximately -25.506 m/s.
Now, to find how far below the pelican the fish is after 2.6 s, we can use the kinematic equation:
s = ut + (1/2)at^2
Where:
s is the distance traveled by the fish
u is the initial velocity of the fish (which is zero since the fish was dropped)
a is the acceleration of the fish due to gravity (-9.81 m/s^2 since gravity pulls objects downward)
t is the time (2.6 s)
Plugging in the values, the equation becomes:
s = 0 * 2.6 + (1/2) * (-9.81) * (2.6)^2
s = -33.6246 m
Therefore, the fish is approximately 33.6246 meters below the pelican after 2.6 s.