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We know that acceleration = dv/dt. If an object is thrown upward, v=0 at the highest point. Hence acceleration (from above relation) must be 0,...

Hello!


Luckily, there is no contradiction. You are right that `a = (dv)/dt,` that `v = 0` at the highest point and that the acceleration due to gravity is constant.


The only incorrect statement you use is that `(dv)/dt = 0` if `v=0.` It is not true in general.


Recall the definition of a derivative `(dv)/dt` or `v',` which are the same. For a function `v(t)`:


`v'(t) = lim_(Delta t->0) (v(t+Delta t) - v(t))/(Delta t).`


If `v(t) = 0` for some specific moment `t,` `v'(t)` becomes:


`v'(t) = lim_(Delta t->0) (v(t+Delta t))/(Delta t).`


But it doesn't have to be zero! For example, in our case `v(t) = V_0-g*t` is a linear function. For any `t,` even for `t = V_0/g,` where `v(t)=0,`


`(v(t+Delta t) - v(t))/(Delta T) = ((V_0-g*(t+Delta t))-(V_0-g*t))/(Delta t) = -g,`


so the limit is also `-g` (this means `+g` downwards).


There is a grain of truth in your assumption, nevertheless. If `v(t)=0` at some entire interval around a point `t_0,` then `(v(t_0+Delta t) - v(t_0))/(Delta t) = 0` for all `Delta t` small enough, and `v'(t_0)=0.`

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