# Why isn't angle of launch used in this kinetic energy equation?

Question: A stick is thrown from a cliff 27 m high with an initial velocity of 18 m/s at an angle of 37 degrees above the horizontal. a) use the law of conservation of energy to determine the speed of the stick just before it hits the ground.

solution:

a busy cat http://i60.tinypic.com/2qmdz74.png

Just wondering why the vertical initial speed vector isn't used? Should 18sin37 be used? I think I am confusing myself with projectile motion.

• Energy is a scalar not a vector, so direction isn't important. It is the magnitude of the velocity vector that is important for calculating energy. You don't have an energy in the x and y directions, you just have energy. – Kenshin Apr 27 '14 at 2:59

So the potential energy plus the kinetic energy at the start equals the kinetic energy at the end. $\frac {1}{2}m{v_i}^2 + mgy = \frac {1}{2}m{v_f}^2$
The energy of the stick depends on its speed which is the magnitude of the velocity vector of the stick. Speed can be expressed as $v_1 = \sqrt{v_{1x}^2+v_{1y}^2}$, where $v_{1x}$ is the horizontal component of the stick's initial velocity vector, $\overrightarrow{v_1}$, and $v_{1y}$ is its vertical component. In this respect, you can see that the answer did in fact depend on the vertical component of the initial velocity vector. Part of the confusion arises from the answer's use of the symbol $v_1$ for speed. I'm sure the author of the question intended for the reader to infer that $v_1$ indicated a speed since it wasn't bolded or arrowed as in $\mathbf{v_1}$ or $\overrightarrow{v_1}$, indicating a vector.