A ball is launched as a projectile with initial speed $v$ at an angle $\theta$ above the horizontal. Using conservation of energy, find the maximum height $h_\text{max}$ of the ball's flight.
Express your answer in terms of $v$, $g$, and $\theta$.
I am doing:
$$\begin{align*}
\frac{1}{2}mv^2 &= mgh_\text{max} + \frac{1}{2}m(v\cos\theta)^2\\
v^2 &= 2gh_\text{max} + (v\cos\theta)^2 \\
h_\text{max} &= \bigl(v^2 – (v\cos\theta)^2\bigr)/2g \\
h_\text{max} &= v^2\bigl(1 – (\cos\theta)^2\bigr)/2g \\
h_\text{max} &= \frac{v^2\sin^2\theta}{2g}
\end{align*}$$
Is that correct, or is there a much easier way…?
Best Answer
Under the constraints of the problem, then yes, what you're doing is correct.
If you weren't required to use conservation of energy, then it would probably be easier to calculate the vertical component of the initial velocity and use 1D kinematics.