A small fish is dropped by a pelican that is rising steadily at 0.61 m/s.The acceleration of gravity is 9.81 m/s2.
How far below the pelican is the fish after the
3.1 s?
Vi=.61
hf-hi=Vi*t-1/2 g t^2 you are looking for the -(hf-hi) where g=9.81, t=3.1
To find how far below the pelican the fish is after 3.1 seconds, we can use the equation of motion for vertical motion:
y = v₀t + 0.5at²
where:
- y is the vertical displacement (distance below the pelican)
- v₀ is the initial velocity (in this case, 0 since the fish is dropped)
- t is the time (3.1 seconds)
- a is the acceleration due to gravity (-9.81 m/s², negative because it acts downwards)
Substituting the values into the equation:
y = (0)(3.1) + 0.5(-9.81)(3.1)²
y = 0 + 0.5(-9.81)(9.61)
y = 0 + (-4.905)(9.61)
y = -47.091 m
Therefore, the fish is 47.091 meters below the pelican after 3.1 seconds.