# Peaked Roof

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1. Aug 24, 2015

### Nishikino Maki

1. The problem statement, all variables and given/known data
A peaked roof is symmetrical and subtends a right angle, as shown. Standing at a height of distance h below the peak, with what initial speed must a ball be thrown so that it just clears the peak and hits the other side of the roof at the same height?

Diagram: http://imgur.com/bi1efMm

2. Relevant equations
$x=vcos(\theta)t$
$y=vsin(\theta)t-\frac{1}{2}gt^2$
$\frac{1}{2}mv^2=mgh$

3. The attempt at a solution
This is supposed to be a kinematics problem, but I wasn't too sure how to do it that way so I used energy principles. Using the above equation, I got that the velocity should be $\sqrt{2gh}$ by simply rearranging the variables, however, the answer key says that it's $\sqrt{5/2gh}$.

2. Aug 24, 2015

### Nathanael

That equation you used is wrong. It assumes the speed is zero when the height is h.

Give it another shot, then explain your thinking.

3. Aug 24, 2015

### Nishikino Maki

Does it have something to do with how the vertical speed is 0 but it is still moving horizontally? In that case the vertical speed at the bottom would be $\sqrt(2gh)$, and the horizontal would be (by working backwards from the answer) $\sqrt(1/2gh)$.

4. Aug 25, 2015

### haruspex

Yes.
That would have dimension 1/speed, so cannot be right.
Knowing the vertical launch speed, how long does it take to reach the top? How far has it moved horizontally in that time?

5. Aug 25, 2015

### Nishikino Maki

Sorry for bad formatting, I meant $\sqrt(1/2) * \sqrt(gh)$.

I decided to do the problem a different way, I used the formula for max height and max range, set max height to half of max range, and solved for the angle. From there I was able to get the max height, which indeed was $\sqrt(5/2)\sqrt(gh)$. Thanks for your help.