# As the speed time graph for a body thrown vertically upwards is not differentiable at the top point so why acc. is taken as 9.8 m/s² at the top point?

In the shown speed - time graph for a body thrown vertically up there is a sharp kink in the graph at the instant for which body is at highest position ie not differentiable so we can't find acceleration. But at this highest point acceleration is due to rate of change of speed only as no angular velocity is there at highest point. Which means no radial acceleration component is there at highest point.

Now my question is using which logic do we say that value of acceleration at highest point is $$9.8 ~\textrm{m/s}^2$$? Does this situation shows that slope of speed - time graph may not represent magnitude of acceleration?

• 9.8m/s/s is for the graph of the component of velocity that points towards earth. The speed is different from that. Apr 6 at 7:24
• @MichaelSeifert sure, good point. I've deleted my comment. Apr 6 at 14:11

The velocity is differentiable at $$v = 0$$ so the acceleration is defined at that point.