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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...?

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I converted your math to LaTeX; since we have the MathJax renderer available on this site, there's no need to use images for math. –  David Z Jul 25 '11 at 3:07

2 Answers 2

up vote 1 down vote accepted

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.

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But in this case using conservation of energy.... –  cMinor Jul 25 '11 at 3:17

I am not sure what David is pointing at in his second paragraph, but you could also consider only the vertical component of the velocity and ignore the horizontal one. The initial vertical velocity is:

$$ v_{vert} = v \sin(\theta) $$


$$ 0.5 m v^2 = m g h $$

this leads to

$$ \begin{align*} 0.5 m (v \sin(\theta))^2 = m g h_{max}\\ \frac{0.5 (v \sin(\theta))^2}{g} = h_{max}\\ h_{max} = \frac{v^2 \sin^2(\theta)}{2 g } \end{align*} $$

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This answer is correct, but the underlying reasoning is likely to be unclear to an introductory student (presumably the target audience here). The second line ($0.5mv^2=mgh$) looks like conservation of energy, but it's not clear why you're allowed to go from there to the next step, where you use just one component of the velocity. In general, conservation of energy doesn't apply to just one component of $v$! If you add a ${1\over 2}v_x^2$ term to both left and right in the third equation, then everything's fine, and since $v_x$ is constant it doesn't change the answer. –  Ted Bunn Jul 25 '11 at 15:07

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