Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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...?

share|improve this question
    
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
add comment

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.

share|improve this answer
    
But in this case using conservation of energy.... –  cMinor Jul 25 '11 at 3:17
add comment

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:

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

Using

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

this leads to

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.