Asked by Anonymous

A small fish is dropped by a pelican that is
rising steadily at 0.60 m/s.
After 2.8 s, what is the velocity of the fish?
The acceleration of gravity is 9.81 m/s2.

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

Answers

Answered by Henry
V=Vo + g*t = -0.60 + 9.81*2.8=26.87 m/s

d = Vo*t + 0.5g*t^2
d = -0.60*2.8 + 4.9*2.8^2 = 36.74 m below the pelican.
Answered by Dan
Your solution is missing the additional distance that the pelican traveled in the 2.8 s.
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions