# If a plane releases a ball flying completely vertically at 300m/s compared to completely horizontally, how will the kinetic energy be different?

If you have a plane flying horizontally at 300m/s , and a ball is released, you can find the final kinetic energy because you know initially it has kinetic energy in the horizontal direction and potential energy via mgh.

my test book says that if the plane were flying vertically the kinetic energy right before it hits the ground would be the same. But if the plane is going 300m/s vertically, then wouldnt the ball rise further than if the plane were dropped when it was going horizontally?

Or is it equal because the horizontal case has horizontal kinetic energy where in the vertical case, that energy is exchanged for vertical?

• Energy is not a vector quantity. You can't have vertical or horizontal kinetic energy. An object just has kinetic energy. – Aaron Stevens Mar 6 '18 at 21:23
• @Aaron's comment is completely correct, but because $\vec{v}^2 = v_x^2 + v_y^2 + v_z^2$ it is possible to deal with a quantity you might call "the contribution to kinetic energy due to motion in the $x$ direction" (and $y$ and $z$ of course) in a manner that is not too silly. Such exercises are useful in making kinetic theory accessible to relatively unsophisticated audiences. But it should be emphasized that $\frac{1}{2}mv_x^2$ is not a "component of kinetic energy". See also: equipartition theorem. – dmckee Mar 7 '18 at 2:29

There's no such thing as horizontal kinetic energy and vertical kinetic energy. Energy is not a vector. However, you have to take into account both horizontal and vertical motion, since the $v$ in $KE = 1/2 mv^2$ is the velocity squared, and velocity is a vector.