What does squaring a vector mean? So,in the 3rd equation of motion,i.e $v²=u²+2as$,what does the square on the final velocity and initial velocity actually mean?And how does it make an effect on their direction?
suppose upwards vector is positive and downwards vector is negative.And,we are throwing a ball from a height of h from ground.In this case,u=-4m/s and there is some final velocity downwards
While using $v²=u²+2as$ here, $u$ is -4m/s and so $u²$ will be 16(m/s)^2. Doesnt this change its direction?
 A: $v^2=u^2-2as$ is usually presented as a formula that applies to one-dimensional motion with constant acceleration $a$ - in this version $u,v,a,s$ are all scalar quantities so multiplying them is not a problem.
However, there is a vector version of this formula which applies to motion in any number of dimensions, where multiplications are replaced by dot products:
$\vec v \cdot \vec v = \vec u \cdot \vec u + 2 \vec a \cdot \vec s$
In this case the acceleration $\vec a$ is assumed to be a constant vector.
A: This formula is elegantly confusing with a deeper hidden meaning.
Explanation of the formula
Given the initial velocity of an object and its acceleration, plus its displacement, we can figure out the velocity of the object at that specific position in all moments of time.
Your confusion
In your case , the final velocity at any given position(or displacement) specified by you, is the same even if you throw the ball upwards (+ve) or downwards (-ve). This is because this formula doesn't include time as a parameter.
Hope it helps!
A: $v^2 = v \cdot v$ - where $\cdot$ indicates the dot product.
The result of the dot product is a scalar, not a vector; it doesn't have a direction.
Example for your vector: $\vec{u} = 0 \vec{i} - 4 \vec{j} + 0 \vec{k}$, because $u$ is completely in the vertical direction and there is no component in the horizontal or perpendicular directions. Then $\vec{u} \cdot \vec{u} = 0 \vec{i} \cdot \vec{i} + 16 \vec{j} \cdot \vec{j} + 0 \vec{k} \cdot \vec{k} = 16$, since the unit vectors dot themselves $= 1$ (make sure you understand why this is the case). 16 is a scalar, it is no longer a vector, as you can see from the fact that it doesn't have $\vec{i}, \vec{j}$, or $\vec{k}$.
A: This equation should be written:
$$
v_{x}^2 = v_{0x}^2 + 2 a_x (x - x_0)\,,
$$
where $v_x = v_x(t) = \vec{v}(t) \cdot \hat{i}$ and $x = x(t) = \vec{r}(t) \cdot \hat{i}$, meaning that it is an equation for the components of the vectors, under the assumption of constant acceleration vector, $\vec{a}$, and a choice of coordinate system.
To see that this is true, recall that the basic equations of motion for a constant acceleration vector are:
\begin{align*}
\vec{r}(t) &= \vec{r}(0) + \vec{v}(t) \,t + \frac{1}{2} \vec{a} t^2\\
\vec{v}(t) &= \vec{v}(0) + \vec{a} t
\end{align*}
which are found by integrating $\frac{{\rm d}^2 \vec{r}}{{\rm d} t^2} = \vec{a}$ twice. What is true for these vector equations is also true for any component of the vectors, under any choice of coordinate system.  So, given such a choice, we have the following equations for the $x$ components of the vectors:
\begin{align*}
x &= x_0 + v_x t + \frac{1}{2} a_x t^2\\
v_x &= v_{0x} + a_x t 
\end{align*}
where I've suppressed the time dependence of $x$ and $v_x$ and written $x_0 = x(0)$ and $v_{0x} = v_x(0)$.
We can do algebra on these two component equations to eliminate time and obtain an equation that relates the $x$ components of velocity and position. Namely, multiplying the first equation by $2 a_x$, we obtain
$$
2 a_x (x - x_0) = 2 a_x v_{0x} t + a_x^2 t^2
$$
and then squaring $(v_x - v_{0x})$ in the second equation and inserting the second equation again:
\begin{align*}
(v_x - v_{0x})^2 &= a_x^2 t^2 \\
v_x^2 - 2 v_x v_{0x} + v_{0x}^2 &= a_x^2 t^2\\
v_x^2 + v_{0x}^2 - 2 (v_{0x} + a_x t) v_{0x} &= a_x^2 t^2\\
v_x^2 - v_{0x}^2 = 2 a_x v_{0x} t + a_x^2 t^2
\end{align*}
and finally inserting our first result into this second result we obtain:
$$
v_x^2 - v_{0x}^2 = 2 a_x (x - x_0)\,.
$$
So the equation you are presenting is actually an equation for the components of the position, velocity, and acceleration vectors along any chosen direction (again, under the assumption that the acceleration vector is constant).  The vector equation version presented above in gandalf61's answer is obtained by summing the $x$, $y$, and $z$ versions of this equation together, but this component equation cannot be determined from that vector equation.
