Question

A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally lets go of the fish, and 1.75 s after the bird lets go, the fish lands in the ocean. (a) Just before reaching the ocean, what is the horizontal component of the fish's velocity in m/s? Ignore air resistance. Assume the bird is initially traveling in the positive x direction. (Indicate the direction with the sigin of your answer.) m/s (b) Just before reaching the ocean, what is the vertical component of the fish's velocity in m/s? Ignore air resistance. Assume upward is the positive y direction and downward is the negative y direction. (Indicate the direction with the sign of your answer.) m/s (e) If the seagull's initial speed were increased, which of the following regarding the fish's velocity upon reaching the ocean would be true? (Select al that apply.) O The horizontal component of the fish's velocity would increase. O The horizontal component of the fish's velocity would decrease. O The horizontal component of the fish's velocity would stay the same. O The vertical component of the fish's velocity would increase. O The vertical component of the fish's velocity would decrease. O The vertical component of the fish's velocity would stay the same.

Answer 1

(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, $g=-9.8 m/s^2$, downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

$v_x = +2.60 m/s$

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

$v_y = u_y +gt$

where

$u_y = 0$ is the initial vertical velocity, which is zero

$g=-9.8 m/s^2$ is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

$v_y = 0+(-9.8)(1.75)=-17.2 m/s$

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.