An apparent paradox in simple kinematics

Question

Among the "Big Five" equations of kinematics is:

V_f² = V_i² + 2 aΔX

Taking this equation at its "face value", it would imply that when Vf = Vi the acceleration has to be 0 or the ΔX is zero. But if we are describing a situation in which an object is thrown on earth from a certain height and the acceleration is g, these two interpretation are mutually contradictory. .

Where is the fallacy ?

Answer 1

I am not sure how a contradiction is arrived at.

Remember that, in projectile motion, the two directions (horizontal and vertical) are independent. So we only deal with one at a time.

In the vertical, if an object is thrown and does not return to the same height, vf does not equal vi. If it returns to the same height, vf = -vi (which is allowed; note that we have to take a square root to solve for either velocity using this equation). But the vertical displacement is also zero.

In the horizontal direction for a projectile, the acceleration is zero. This is because, by definition, the only allowed acceleration for a projectile is the acceleration of gravity, which is in the vertical, and thus has no bearing on the horizontal direction. As a result, the horizontal velocity of a projectile is constant over its entire flight as a projectile.

If there is a contradiction I am missing, please let me know, and we can talk about it further.

Also, I usually see it and teach it as only four kinematic equations. There are five quantities, as I usually present, but only four universally applicable equations, assuming acceleration is constant.

Answer 2

A partial answer, at least, can be had by taking into account the vector character of velocities and acceleration.

The vector form of the equation is:

Vf dot Vf = Vi dot Vi + 2 g dot DX

Here dot means the vector dot product. So if g is the vertical direction and DX is in the horizontal direction, the dot product of g and DX is zero. With this situation Vf dot Vf = Vi dot Vi and thus Vf = Vi