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A ball is thrown straight up from the ground. How high will it go?

The height of the ball at time t\, is x(t)\,, and the force acting on the ball is -mg\,.

Newton's second law gives the relation

m{\frac  {d^{2}x}{dt^{2}}}+mg=0\,

Let m=1\, for convenience.

{\frac  {d^{2}x}{dt^{2}}}+g=0\,

Integrate the last equation.

{\frac  {dx(t)}{dt}}+gt+C=0\,

To get C\,, start with an initial velocity v_{0}\, and set t=0\,


So now,

{\frac  {dx}{dt}}+gt-v_{0}=0\,

Integrate again.

x(t)+{\frac  {1}{2}}gt^{2}-v_{0}t+D=0\,

Since we have set x(0)=0\,, D=0\,

The equation of position is

x(t)=-{\frac  {1}{2}}gt^{2}+v_{0}t\,

To find the maximum, set the first derivative equal to 0.


t_{{max}}={\frac  {v_{0}}{g}}\,

Check that the second derivative at this point is negative.

x''\left({\frac  {v_{0}}{g}}\right)=-g\,

So the answer is this point t_{{max}}\, evaluated in the position function.

x_{{max}}=x\left({\frac  {v_{0}}{g}}\right)={\frac  {v_{0}^{2}}{2g}}\,

Ordinary Differential Equations