Solving a Kinematics Problem: Throwing a Ball to Clear a Peaked Roof

[Nishikino Maki]

Homework Statement

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

Homework Equations

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

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}.

 

[Nathanael]

... I used energy principles. Using the above equation, I got that the velocity should be \sqrt{2gh} ...

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.

 

[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).

 

[haruspex]

the vertical speed is 0 but it is still moving horizontally?

Yes.

and the horizontal would be (by working backwards from the answer) $\sqrt{\frac 1{2gh}}$.

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?

 

[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.