Why is Acceleration Non-Zero at Maximum Height?

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Acceleration remains non-zero at maximum height because it is influenced by gravitational force, which is constant at -9.8 m/s², regardless of the object's velocity. When an object is thrown upwards, its velocity reaches zero at the peak, but the acceleration due to gravity continues to act on it, pulling it back down. The confusion arises from misunderstanding that the derivative of velocity (acceleration) does not depend solely on the instantaneous velocity but rather on the forces acting on the object. The discussion highlights the importance of differentiating velocity as a function of time to understand the behavior of the object throughout its motion. Thus, acceleration is always present, even when velocity is zero at the maximum height.
nebbione
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Hi everyone! If i know that acceleration is the derivative of the velocity, why when i throw an object in the air when the object reaches the maximum height its velocity is 0 m/s and its acceleration is still -9.8 m/s^2 ? I mean, the derivative of 0 m/s is not -g but still 0.

Who can explain me this ?
 
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What is the derivative of y = 4x - 4? What is the derivative of that when x = 1?
 
the derivative is 4 and when x=1 is still 4.
 
Do you see your mistake now?
 
Velocity in this case(magnitude) is not constant.Therefore you have to differentiate velocity as a function of time this and find its value at the required time.However,since velocity increases uniformly and I've is linear,its derivative will be constant
 
Vanadium 50 -

Clever - interesting how you explained what he did wrong
without explicitly stating what he did wrong. That is a good
strategy to employ when teaching others. Thank You.
 

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