# 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

## 2 Answers

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.

You're right that if the plane is flying vertically then the ball will rise further than if the plane were flying horizontally. However if the plane were flying vertically, it isn't moving in the horizontal direction. In that case the velocity of the ball right before it hits the ground is completely in the vertical direction. In the other case, the ball has a (smaller) velocity in the vertical direction and a velocity in the horizontal direction. Before you can use the kinetic energy equation, you need to vector-sum these two velocities.

If you work through the math, you should find that the two velocities have the same magnitude (neglecting air resistance). A shortcut is that in the latter case, when the ball hits the ground, its horizontal velocity is still 300m/s. You can then compute its vertical velocity, vector sum the two, and compare with the first case.

Neglecting air resistance, in the horizontal case the ball has kinetic energy T when it is released and potential energy V equal to mgh. As it falls it’s potential energy is converted to kinetic energy so the total kenetic energy when it hits the ground is T+V. In the vertical case, the ball with kinetic energy T rises, slows, then stops. At this point all of its energy has been converted to potential energy. It then falls so that when it passes the point where it was released it has regained exactly the same kinetic energy it had when it was released. From there V=mgh is again converted to kinetic energy so that when it hits the ground it’s kinetic energy is again T+V.