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?
Just use your good old equation of motion
s = Vot + 1/2 at^2
s = .61t - 4.9t^2
Of course, that's how far the fish drops. You have to add the height the bird has gained in the 3.1s.
To find the distance below the pelican at a given time, we can use the equations of motion.
We'll start by assuming that the initial height of the fish (above the ground) is zero.
1. First, we need to find the initial velocity of the fish when it is dropped by the pelican. The pelican is rising steadily at a velocity of 0.61 m/s, so the fish will also have this velocity initially when dropped.
2. Next, we need to find the fish's vertical displacement after 3.1 seconds. We can use the equation:
s = ut + (1/2)gt^2
where:
- s is the vertical displacement (distance below the pelican)
- u is the initial velocity (0.61 m/s)
- t is the time (3.1 s)
- g is the acceleration due to gravity (-9.81 m/s^2, negative because it acts downward)
Plugging in the values:
s = (0.61 * 3.1) + (0.5 * -9.81 * (3.1)^2)
3. Calculate the result:
s = (1.891) + (-45.266)
s = -43.375 meters
Therefore, the fish is approximately 43.375 meters below the pelican after 3.1 seconds.