Question

005 (part 1 of 2) 10.0 points

Answer 1

Answer:

7.0 m/s

Explanation:

First of all, we need to find the time it takes for the fish to reach the water below. This can be done by considering the vertical motion of the fish only, which is a free fall motion, so by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s = 4.6 m is the vertical displacement of the fish (choosing downward as positive direction)

u = 0 is the initial vertical velocity of the fish

t is the time

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

Solving for t, we find:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(4.6)}{9.8}}=0.97 s$

Now can consider the horizontal motion of the fish; since there are no forces along this direction, the fish travels at constant horizontal velocity, and so the distance travelled is

$d=v_x t$

Here we have

d = 6.8 m is the horizontal distance travelled by the fish

t = 0.97 s is the time of flight

Solving,

$v_x = \frac{d}{t}=\frac{6.8}{0.97}=7.0 m/s$

And since the horizontal velocity of the fish is constant, and it is equal to the initial velocity of the pelican, this means that the initial speed of the pelican was 7.0 m/s.