# Interpretation of 2 opposite values of velocity in one dimensional motion

Suppose an object is thrown upward with initial throw velocity $V_i$ and object's velocity become zero ($V_f=0$) as it arrives at a height $h$ from the throw point. The motion is along y-axis and the upward direction is taken positive. Then we have following equation to calculate Vi.

$$-2gh=V_f^2 - V_i^2$$

Which implies,

$$V_i = \pm \sqrt{2gh+V_f^2}$$

Now we have 2 values for initial velocity and both satisfies the equation however only positive value is the true one because initial velocity got to be in the upward direction which we assumed positive. So what is the negative value telling us considering that it is not an extraneous root?

This equation of motion you are using i.e. $V_f^2 - V_i^2 = 2as$ is a scalar equation. Hence the solution, $$|V_i| = \sqrt{V_f^2 - 2as}$$ tells that the absolute value of initial velocity i.e. initial speed is $\sqrt{V_f^2 - 2as}$ . Since speed is rate of change of distance with time, and distance cannot be -ve as it is the physical length covered by an object in its motion, hence the speed is also +ve.