How Does Gravity Affect a Dropped Fish's Motion?

[arizona_cards_11]

Question:

54) A small fish is dropped by a pelican that is rising steadily at 0.50 m/s.

a) After 2.5 s, what is the velocity of the fish? Ans: -24 m/s

b) How far below the pelican is the fish after 2.5 s? Ans: 31m

My Work/Question:

The fish would be moving at the same velocity as the pelican, which equals the (Vi). The Accel would equal -9.81 m/s^2 due to gravity.

I used Vf = (Vi) + (A)(T) to find the velocity of the fish.

My question on B: Would I use the... delt X = (Vi)(t) + (1/2)(A)(T)^2...formula for B?

 

[arizona_cards_11]

DO NOT REPLY...I've already figured it out

 

[kyle8921]

Yes, you would use the displacement formula, Δx = (Vi)(t) + (1/2)(a)(t)^2, to find the distance the fish has fallen after 2.5 seconds. This formula takes into account the initial velocity, the acceleration due to gravity, and the time elapsed. Plugging in the given values, we get Δx = (0.50 m/s)(2.5 s) + (1/2)(-9.81 m/s^2)(2.5 s)^2 = 31.06 m. Therefore, after 2.5 seconds, the fish would be 31.06 meters below the pelican.