# Does the kinetic energy change depending on the direction of the velocity? [closed]

The kinetic energy is calculated with the following formula: $$E_k = \frac 12 mv^2$$ Does $v$ represent the value of the velocity? I mean, if you have a specific direction of the velocity is the formula: $$E_k = \frac 12 mv_x^2$$ Or $$E_k = \frac 12 m\left(\sqrt{v_x^2 + v_y^2}\right)^2$$

No matter which of the equations are the correct one, the direction will influence in which is the highest height a body can reach.

Having said that $$E = E_k + U_G \;\;\;\text{(Mechanical energy)}$$ How would I calculate the highest height a body can reach? I know how to calculate it when I know that it is just vertical movement.

## closed as off-topic by ACuriousMind♦, user36790, Kyle Kanos, Sebastian Riese, Bill NFeb 23 '16 at 4:16

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• $v$ is the magnitude of $\boldsymbol v$, i.e., $v^2\equiv v_x^2+v_y^2+v_z^2$. Note that this quadratic form is invariant under rotations, which means $v^2$ is independent of the direction of $\boldsymbol v$. – AccidentalFourierTransform Feb 20 '16 at 17:51
• @AccidentalFourierTransform I didn't understand how can you see it is independent. – Pichi Wuana Feb 20 '16 at 17:52
• well, $v^2\equiv \boldsymbol v\cdot\boldsymbol v$, i.e., it is the scalar product of $\boldsymbol v$ with itself. Therefore, $v^2$ is a scalar, which means it is rotational invariant (see scalar on wikipedia) – AccidentalFourierTransform Feb 20 '16 at 17:55
• @AccidentalFourierTransform From my knowledge I know that direction will change the highest height. Even in my book, a exercise tells us that two projectiles were launched from same place with same magnitude of velocity, but with different directions. They took two different parabola paths. I needed to find if the place where the parabolas are cut, they will meet each other... – Pichi Wuana Feb 20 '16 at 18:02

Say we fire a projectile from the $xy$-plane: The $xy$-plane is the horizontal plane, the $z$-axis is the Normal, running through the centre of the Earth.

At launch ($t=0$) the projectile has velocity vectors $\vec{v_x}$, $\vec{v_y}$ and $\vec{v_z}$. The kinetic energy $K$ is indeed given by:

$$K=\frac12m(v_x^2+v_y^2+v_z^2)$$

So what height would the projectile reach (ignoring of course air drag)?

Understand that only one force acts of the projectile and it acts in the (minus) $z$-direction: gravity.

As the projectile gains height, it also gains potential energy $U$:

$$U=mgz$$

Because gravity only acts in the $z$ direction, only $\vec{v_z}$ is affected and not $\vec{v_x}$ or $\vec{v_y}$. The latter two vectors will influence how far the projectile lands from the launch point but not what height it will reach.

For that reason we can write:

$$mgz=\frac12mv_z^2,$$

from which the maximum $z$ is calculated.

Note that energy conservation is respected. We start off with:

$$T=K=\frac12m(v_x^2+v_y^2+v_z^2)$$

And at the highest point:

$$T=U+K=mgz+\frac12m(v_x^2+v_y^2),$$

with:

$$mgz=\frac12mv_z^2$$