Answer to Question #79525 in Physics for rachel

Question

A small fish is dropped by a pelican that is rising steadily o.50m/s, after 2.5 s, what is the velocity below the pelican is the fish of the fish?

Answer 1

(a) We can find the velocity of the fish from the kinematic equation:
v=v_0-gt,
here, v_0 is the initial velocity of the fish, g=9.8 m⁄s^2 is the acceleration due to gravity and t is the time.
Then, we get:
v=v_0-gt=0.50 m/s-9.8 m/s^2 ∙2.5 s=-24 m/s.
The sign minus indicates that the velocity of the fish directed downward.
(b) Let’s first find the distance that the fish fall after being dropped by a pelican (we choose the upwards as the positive direction):
d_1=v_0 t-1/2 gt^2=0.50 m/s∙2.5 s-1/2∙9.8 m/s^2 ∙(2.5 s)^2=-29.4 m.
Then, we can find the distance by which the pelican rise during that time:
d_2=vt=0.5 m/s∙2.5 s=1.25 m.
Finally, we can find the distance between the pelican and the fish:
d=d_1+d_2=29.4 m+1.25 m=31 m.