A small fish is dropped by a pelican that is rising steadily at 0.22 m/s.?

Question

After 3.3 s, what is the velocity of the fish?

and

How far below the pelican is the fish after 3.3 s?

show me how to solve it plz!

Answer 1

The pelican and the fish were rising up with a velocity of 0.22 m/s just before the instant it was dropped.
So initial velocity of the fish is 0.22 m/s upwards.
Accl. is g(=9.8m/s^2) downwards.
Use equations of motion in a straight line to solve.

v=u+at
Considering upward to be negative,
v= -0.22+9.8*.33

In .33 seconds, the fish falls by
s= -0.22*0.33+0.5*9.8*.33^2 m
Pelican rises by 0.22*0.33 m
Add the distances to get the answer