What is the Acceleration of a Dropped Shell?

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The discussion centers on calculating the acceleration of a shell dropped by a glaucous-winged gull at 11.5 meters above the ground. Initially, the shell has zero acceleration while the bird ascends at a constant speed. Upon release, the shell enters free fall, experiencing an acceleration of 9.81 m/s² due to gravity. The initial attempt to use the equation v² = v(i) + 2a(x - x(0)) was incorrect, as it did not account for the shell's immediate transition to free fall. For finding the maximum height, participants suggest using constant acceleration formulas that do not require time.
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Homework Statement



A glaucous-winged gull, ascending straight upward at 5.10 m/s drops a shell when it is 11.5m above the ground. What is the magnitude of the shell's acceleration just after it is released?

Homework Equations


The only equation that I can think of would be v^2= v(i) + 2a (x-x(0)) because I have no time value

The Attempt at a Solution


Using this equation I got an answer of 1.13 m/s^2, but on my hw, it claims it's incorrect. So help please!
 
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This sounds like one of those questions where the book is trying to trick you into thinking the problem needs more work than is necessary.

Before the bird let's go of the shell it has zero acceleration since it is moving up at a constant speed with the bird. When the bird let's go of the shell it enters free fall. That should give you your answer.
 
Thanks, you were right. But question, how am I supposed to find the maximum height? Apparently it's not 11.5m, so how would I start approaching the problem?
 
Use the constant acceleration formulas. In fact there is one formula that relates all the variables you have to part of the answer you are looking for; it doesn't involve time.
 
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