Vertical Shot

In a more general approach, it is assumed now that the object does not only fall to the ground but instead moves upwards with the initial speed $v_0$. This could be for instance a stone that is thrown in the air by a person. In this case, the equation of motion has to be modified according to $$y(t) = -\frac{1}{2}gt^2 + v_0 t + h$$ In order to compute the time from the moment of release until the object reaches the ground, the following equation has then to be solved: $$-\frac{1}{2}gt^2 + v_0 t + h = 0$$ This results in $$\boxed{t = \frac{v_0}{g}\pm \sqrt{\frac{v_0^2}{g^2}-\frac{2h}{v_0}}}$$ The overall time depends on the initial speed $v_0$ of the object and the height $h$ from which it was thrown. It is also possible to calculate the time at which the object reaches the highest at which the speed, and therefore the derivative with respect to time, gets 0: $$\frac{\mathrm{d}y}{\mathrm{d}t} = -gt + v_0 = 0$$ The solution is then given as $$\boxed{t_\mathrm{max} = \frac{v_0}{g}}$$ Inserting this value back into the original function makes it also possible to calculate the maximum height: $$\boxed{y_\mathrm{max} = -\frac{1}{2}\frac{v_0}{g^2} + \frac{v_0^2}{g} + h}$$