I'm having some difficulty understanding how to find the maximum height using conservation of energy.
This is the picture I'm looking at:
and this is how you find it:
$$\begin{align*}
\frac{1}{2}mv^2 &= mgh_\text{max} + \frac{1}{2}m(v\cos\theta)^2\\
v^2 &= 2gh_\text{max} + (v\cos\theta)^2 \\
h_\text{max} &= \bigl(v^2 – (v\cos\theta)^2\bigr)/2g \\
h_\text{max} &= v^2\bigl(1 – (\cos\theta)^2\bigr)/2g \\
h_\text{max} &= \frac{v^2\sin^2\theta}{2g}
\end{align*}$$
However, I am confused about a few things. I know that all those equations stem from using $K_{i} + U_{i} = K_{f} + U_{f}$. The initial potential energy is 0 because it just started moving, correct? How come we needed to use the x-component of the kinetic energy to use $K_{f}$ (I assume that's where the cos came from) and not for $K_{i}$, where it's just $1/2mv^2$. I don't understand the importance of it?
Best Answer
The initial potential energy is zero because the ball starts off at essentially ground level, and potential energy is being defined as being zero at ground level.
The initial velocity is a vector of magnitude v that points up at an angle $\theta$ from the ground. The components of that initial velocity are $v_x(0)=v \cos\theta$ in the horizontal direction, and $v_y(0)=v \sin\theta$ in the vertical direction.
$v_y(t)$ changes with time due to gravity, with $v_y(t_{apex})=0$ when the ball is at its apex.
$v_x(t)$ doesn't change with time during the ball's path, because there is no horizontal force on the ball. Since at the ball's apex, $v_y(t_{apex})=0$ and $v_x$ is still given by $v_x(t_{apex})=v \cos \theta$, the ball's speed at the apex is $v \cos\theta$, which is why that speed is used for the ball's speed in the expression for the kinetic energy of the ball at its apex.