# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Community, Kyle Kanos, Sebastian Riese, Bill N
If this question can be reworded to fit the rules in the help center, please edit the question.

• $v$ 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:

$$mgz=\frac12mv_z^2$$